8
Work and Energy
1.
The mass of cyclist with the bike is 90 kg. Calculate the increase in kinetic energy if the speed increases from 6.0 km/h to 12 km/h.
Given that
⇒ 2.
(
)
.
/
. /
A block of mass 2.00 kg moving at a speed of 10.00 m/s accelerates at Compute its final kinetic energy.
for 5.00 s.
With usual notation we are given that
( )
⇒ ⇒ 3.
A box is pushed through 4.0 m across a floor offering 100 N resistance. How much work is done by the resisting force ?
Given that ( ) 4.
A block of mass 5.0 kg slides down an incline of inclination 30° and length 10 m. Find the work done by the force of gravity.
Given that (
⇒ 5.
)(
sin
)
A constant force of 2.50 N accelerates a stationary particle of mass 15 g through displacement of 2.50 m. Find the work done and the average power delivered.
The given data is ( ) x
⇒ ⇒
.
/
8
Work and Energy √ .
√ ⇒ ⇒ 6.
(
P=
/(
)
)
A particle moves from a point ( during which a certain force particle during the displacement.
( )
( ) to another point ( ) ( ) ) acts on it. Find the work done by the on the
) (
From the given data (
⇒ 7.
)(
)
A man moves on a straight horizontal road with a block of mass 2 kg in his hand. If he covers a distance of 40 m with an acceleration of . Find the work done by the man on the block during the motion.
Here ( ⇒ 8.
) )
A force acts on a particle in the x-direction where a and b are constants. Find work done by this force during a displacement from .
Here ∴ 9.
(
∫ (
( )
) 0
(
) 1
A block of mass 250 g slides down an incline of inclination 37° with a uniform speed. Find work done against the friction as the block slides through 1.0 m.
Let is the frictional force. As the block moves down the incline with uniform speed the acceleration and consequently the net force acting on the block along the incline are zero. But gravitational force along the incline is sin where m is the mass of the block. ∴ sin ⇒ sin To find work done against friction we note that the displacement and the force against the friction have same direction i.e. down the incline. ( )( ) . / ( ) ∴ cos sin
8
Work and Energy
10. m M
F
A block of mass m is kept over an other block of mass M and the system rests on a horizontal surface. A constant horizontal Force F acting on the lower block produces an acceleration in the system, the two blocks always move together. (a) Find the coefficient of friction ( ) between the bigger block and the surface. (b) Find the frictional force acting on the smaller block. (c) Find the work done by the force of friction on the smaller block by the bigger block during the displacement d of the system.
a)
Let be the coefficient of friction between the bigger block and the surface and that between the bigger block and the smaller block. Let be the frictional forces between bigger block and surface and the bigger block and smaller block respectively. Then the forces on the bigger block in the horizontal direction are F in the forward direction and in reverse direction. ( ) ( ) ( ) ⇒ ( ) ⇒
(
)
⇒ b)
⇒ c)
(
)
Since the only force acting on the smaller block is the frictional force we have (
)
From part (b) we get (
)
11. A box weighing 2000 N is to be slowly slid through 20 m on a straight track having friction coefficient 0.2 with the box. (a) Find the work done by the person pulling the box with a chain at an angle θ with the horizontal (b) Find the work done when the person has chosen the value of θ which ensures him the minimum magnitude of the force
a)
Let F be the force applied by the person at the angle θ with horizontal Let N be the normal force of the track on the box. Then sin ( sin ) ⇒ cos (
b)
)(
)(
)
⇒ The force will be minimum when cos
sin
is maximum. Equating its first
8
Work and Energy derivative to zero we have sin cos ⇒ tan ⇒
12. A block of weight 100 N is slowly slid up on a smooth incline of inclination 37° by a person. Calculate the work done by the person in moving the block through a distance of 2.0 m, if the driving force is (a) parallel to the incline and (b) in the horizontal direction.
a)
Let F be the force applied by the person parallel to the incline. Since acceleration and hence net force is zero, we get sin cos . /( ) b)
In this case the force F is applied in the horizontal direction but the displacement is along the incline i.e. at an angle 37° with the horizontal. Resolving the forces F and mg along the incline and noting that acceleration is zero, we get cos sin ⇒ tan ( ). / ⇒ cos sin
13. Find the average frictional force needed to stop a car weighing 500 kg in a distance of 25 m if the initial speed is 72 km/h.
Let
be the required average frictional force. Then the acceleration is
Also Using the formula ⇒ . ⇒
we get /
14. Find the average force needed to accelerate a car weighing 500 kg from rest to 72 km/h in a distance of 25 m.
Let
be the required average force. Then the acceleration is
Also Using the formula ⇒ .
we get /
8
Work and Energy ⇒
15. A particle of mass moves in a straight line with its velocity varying with the distance travelled according to the equation √ , where is constant. Find total work done by all the forces during a displacement from .
The acceleration is given by . / a√ x √
a
The force therefore is a constant force given by
.
⇒ 16. A block of mass 2 kg kept at rest on an inclined plane of inclination 37° is pulled up the plane by applying a constant force of 20 N parallel to the incline. The force acts for one second. (a) Show that the work done by the applied force does not exceed 40 J. (b) Find the work done by the force of gravity in that one second if the work done by the applied force is 40 J. (c) Find the kinetic energy of the block at the instant the force ceases to act. Take .
(a) Let be the applied force and be the acceleration of the block along the incline. Resolving the acceleration due to the force of gravity along the incline, we see this component of acceleration is –10 sin 37°. Thus we have sin ∴ If, however, the surface of the incline is not smooth, the friction will retard the motion of the block. Thus maximum possible displacement is . If is the work done by the applied force, we have ( ) (b) The height through which the block has gone up in one second is given by sin Since the displacement is in opposite direction to force of gravity, the work done by gravity is ( )( ) (c) If be the velocity after one second, we get by using data from part (a) If K be the kinetic energy after the force ceases to act, we have 17. A block of mass 2.0 kg is pushed down an incline of inclination 37° with a force of 20 N acting parallel to the incline. It is found that the block moves on the incline with an acceleration of . If the block started from rest, find the work done (a) by the applied force in the first second. (b) by the weight of the block in the first second. (c) by the frictional force acting on the block in the first second. Take .
(a) Here the forces are the applied force parallel to the incline acting downwards, the force of gravity and the frictional force . If s is the distance
8
Work and Energy covered by the block along the incline then ∴ (b) Work done by the weight is the work done by the force of gravity. ( )( ) . / ∴ sin (c) From Newton's second law we have sin ( ). / ( )
⇒ ∴
18. A 250 g block slides on a rough horizontal table. Find the work done by the frictional force in bringing the block to rest if it is initially moving at a speed of 40 cm/s if the friction coefficient between the table and the block is 0.1. How far does the block move before coming to rest?
If is the frictional force and N the normal force exerted by the table on the block we have . /( ) ∴ Putting
. (
and )
/ in
we get
⇒ Since the displacement is in opposite direction to the direction of the frictional force we have ( ) 19. Water falling from a 50 m high fall is to be used for generating electric energy. If kg of water falls per hour and half the gravitational potential energy can be converted into electric energy, how many 100 W lamps can be lit ?
The mass m of the water which falls in one second is The potential energy of the mass of water is ( )( ) Since half of this potential energy gets transformed into electric energy we have Thus 122 lamps of 100 W can be lit with the electric energy produced. 20. A person is painting his house walls. He stands on a ladder with a bucket containing paint in one hand and a brush in the other. Suddenly the bucket slips from his hand and falls down on the floor. If the bucket with the paint had mass of 6.0 kg and was at a height of 2.0 m at the time it slipped, how much gravitational potential energy is lost together with the paint?
From the given data the loss of potential energy is given by ( )( )
8
Work and Energy
21. A projectile is fired from the top of a 40 m high cliff with an initial speed of 50 m/s at an unknown angle. Find its speed when it hits the ground.
From the law of conservation of energy we have ⇒ √(
⇒
(
)(
))
√
22. The 200 m free-style women's swimming gold medal at the Seoul Olympics 1988 went to Heike Friendrich of East Germany when she set a new Olympic record of 1 minute and 57.56 seconds. Assume that she covered most of the distance with uniform speed and had to exert 460 W to maintain her speed. Calculate the average force of resistance offered by the water during the swim.
Let be the average force of resistance of water. Then work done by distance of 200 m is
in covering the
This work is done in 1 minute and 57.56 seconds at a uniform rate. Thus rate of work done per second, which is also the Power exerted by the athlete given to be 460 W, is . ⇒ (
⇒
)
23. The US athlete Florence Griffith Joyner won the 100 m sprint gold medal at Seoul Olympics 1988 setting a new Olympic record of 10.54 s. Assume that she achieved her maximum speed in a very short time and then ran the race with that speed till she crossed the line. Take her mass to be 50 kg. (a) Calculate the kinetic energy of Griffith Joyner at her full speed. (b) Assuming that the track, the wind etc. offered an average resistance of one tenth of her weight, calculate the work done by the resistance during the run. (c) What power Griffith Joyner had to exert to maintain uniform speed?
(a) Since the race was run at a uniform velocity, we have ∴ (b) Let
(c)
( )( ) be the force of resistance then ( )
∴ Since the speed is uniform the acceleration is zero and the net force is also zero. Thus the force F exerted by the athlete on the track is equal in magnitude (49 N) and opposite in direction to that of force of resistance. ∴ ⇒
8
Work and Energy
24. A water pump lifts water from a level 10 m below the ground. Water is pumped at a rate of 30 kg/minute with negligible velocity. Calculate the minimum horse power the engine should have to do this?
The work done by the pump in one second is . / ( )( ) ∴
x
25. An unruly demonstrator lifts a stone of mass 200 g from the ground and throws it at his opponent. At the time of projection, the stone is 150 cm above the ground and has a speed of 3.00 m/s. Calculate the work done by the demonstrator during the process. If it takes one second for the demonstrator to lift the stone and throw, what horse power does he use?
Here Using the formula (
we get
)
( ) ⇒ Here is the net force acting on the stone i.e. the force demonstrator minus the force of gravity. We, therefore have ( ) ( )-( ⇒ , Since it takes one second for the lift, the power is x
applied by the
)
26. In a factory it is desired to lift 2000 kg of metal through a distance of 12 m in one minute. Find the minimum horse power of the engine to be used?
Since it takes 60 seconds to lift the 2000 kg metal to 12 m height, the acceleration is small and therefore the force to be applied by engine is . ( ) ∴ This work is done in 60 seconds. ⇒ 27. A scooter company gives the following specifications about its product. Weight of the scooter = 95 kg. Maximum speed = 60 km/h. Maximum engine power = 3.5 HP. Pickup time to get the maximum speed – 5 s. Check the validity of these specifications.
Here
Now ⇒
gives us
8
Work and Energy ⇒ . / ⇒ Thus the scooter will require maximum engine power of 7 HP to reach 60 km/h speed in 5 seconds and therefore the claim seems overstated.
28. A block of mass 30.0 kg is being brought down by a chain. If the block acquires a speed 40.0 cm/s in dropping down 2.00 m, find the work done by the chain during the process.
From the given data we have The acceleration
due to the force applied by chain is
As value of g has been taken positive chain is ( )( )
has upward direction. Thus work W done by
29. The heavier block in an Atwood machine has a mass twice that of the lighter one. The tension in the string is 16.0 N when the system is set into motion. Find the decrease in the gravitational potential energy during the first second after the system is released from rest.
Let masses of the two blocks be and . Let be the acceleration of the heavier block. Then the acceleration of the lighter block is – . The force forces on the two blocks being tension and force of gravity, we have …I … II Subtracting twice II from I we have ⇒ If is the height lost by the heavier block then Let and be the changes in gravitational potential energy of the heavier and lighter blocks. Then loss of potential energy of the system is ( ). / 30. The two blocks in an Atwood machine have masses 2.0 kg and 3.0 kg. Find the work done by gravity during the 4th second after the system is released from rest.
Let be the acceleration of the lighter block. Then the acceleration of the heavier block is – . The force forces on the two blocks being tension and force of gravity, we have …I … II ⇒ Let be the velocity of the lighter block after third second. Then
8
Work and Energy
If
is the height gained by the lighter block during 4th second, then
∴ 31. Consider the situation shown in figure. The system is released from rest and the block of mass 1.0 kg is found to have a speed 0.3 m/s after it has descended through a distance of 1.0 m. Find the coefficient of kinetic friction between the block and the table. T 4.0 kg
T T 2T 1 kg
Let μ be the coefficient of kinetic friction between the block and the table Let T be the friction on the upper string. Then 2T is the tension in the lower string. Let be the acceleration of the 1 kg block. Then …I The force on the 4 kg block are the frictional force and T. Therefore … II Also from the given data and ( ) we have ⇒ … III Solving I II and III for μ we get ⇒ 32. A block of mass 100 g is moved with a speed of 5.0 m/s at the highest point in a closed circular tube of radius 10 cm kept in a vertical plane. The cross section of the tube is such that the block just fits in it. The block makes several oscillations inside the tube and finally stops at the lowest point. Find the work done by the tube on the block during the process.
The work done by the tube on the block is equal to change in kinetic and potential energies or K U But ( ) We take potential energy at point B as zero. Thus ( )( ) 33. A car weighing 1400 kg is moving a speed of 54 km/h up a hill when the motor stops. If it is just able to reach the destination which is at a height of 10 m above the point, calculate the work done against friction (negative of the work done by the friction).
8
Work and Energy Since the car is moving when the motor stops, it has kinetic energy. Thus
⇒ Let gravitational potential energy be zero at the starting point. Then the potential energy at the terminal point is ( )( ) Let W be the work done against friction during ascent. Then –W is the work done by the frictional force. ⇒ ⇒ 34. A small block of mass 200 g is kept at the top of a frictionless incline which is 10 m long and 3.2 m high. How much work was required (a) to lift the block from the ground and put it at the top? (b) to slide the block up the incline? What will be the speed of the block when it reaches the ground if (c) it falls off the incline and drops vertically on the ground (d) it slides down the incline ? Take .
(a) In this case the work done is the negative of the work done by the gravitational force. ( ) ( )( ) ∴ (b) Let the block be slowly slid up the incline. Since the net acceleration is zero the force required to slide it up the incline is ( ) sin ( ) ∴ (c) If the block falls its acceleration is and initial velocity is zero. ∴ ⇒ √ (d) By work – kinetic energy theorem Here
is the work done on the block (by gravitational force)
⇒ ⇒
√
(
)
35. In a children's park, there is a slide which has a total length of 10 m and a height of 8.0 m. Vertical ladder is provided to reach the top. A boy weighing 200 N climbs up the ladder to the top of the slide and slides down to the ground. The average friction offered by the slide is 3/10th of his weight. Find (a) the work done by the ladder on the boy as he goes up, (b) the work done by the slide on the boy as he comes down. Neglect any work done by forces inside the body of the boy.
(a) As the boy pushes down on the ladder, by Newton's third law, the ladder pushes the boy up. Since the boy goes up the ladder slowly, the change in kinetic energy
8
Work and Energy is zero and the work done on the boy by the ladder is equal to work done by the gravitational force on him but in opposite direction. ( ) ( ) ∴ (b) As the boy slides down, a frictional force ( ) does work the boy. Since force and displacement are in opposite directions, we have ( )
on
36. Fig shows a particle sliding on a frictionless track which terminates in a straight horizontal section. If the particle starts slipping from the point A, how far away from the track will the particle hit the ground?
From the work – kinetic energy theorem, we get The work done is U and the initial velocity of the particle is zero ∴ ⇒ √ because h=1.0–0.5=0.5 m √ Now the particle is projected with this velocity √g parallel to x-axis (i e with θ from a height 0.5 m. From the equations of projectile motion we have ( sin ) ( )
⇒ ⇒ ⇒
)
√ √
( cos )
√
37. A block weighing 10 N travels down a smooth curved track AB joined to a rough horizontal surface. The rough surface has friction coefficient of 0.20 with the block. If the block starts slipping on the track from a point 1.0 m above the horizontal surface, how far will it move on the rough surface?
Then the work done by the gravitational force is As the block moves down some of the potential energy of the block is converted into kinetic energy. The block keeps moving till the frictional force between the rough surface and the block completely offsets the kinetic energy of the block. Therefore after the start of the block's journey on the rough surface √ √ Also Putting the values in
we get
( )
38. A uniform chain of mass
and length overhangs a table with its 2/3rd part on the table. Find
8
Work and Energy the work to be done by a person to put the hanging part back on the table.
One third part of the chain is hanging from the table. A small element of the chain's hanging part distance below the table will require work to be done, where . / To find work done W let us integrate it between 0 and l/3 ∴
∫ 0 1
39. A uniform chain of length L and mass M overhangs a horizontal table with its 2/3 rd part on the table. The friction coefficient between the table and the chain is μ Find the work done by the friction during the period the chain slips off the table.
Let us consider an element of the chain on the table distance from the edge of the table. As this element is made to fall off the table the work done by the frictional force is . / Since two third part of the chain is on the table the integration for finding the work to be done by the frictional is between 0 and 2L/3 and noting that the frictional force and displacement are in opposite direction we have ⇒
∫ 0
1
40. A block of mass 1 kg is placed at the point A of a rough track shown the figure. If slightly pushed towards right, it stops at the point B of the track. Calculate the work done by the frictional force on the block during its transit from A to B.
Since the change in kinetic energy is zero, the total work done by the frictional force is negative of the work done by gravitational force. ( ) ( ) ∴ 41. A block of mass 5.0 kg is suspended from the end of a vertical spring which is stretched by 10 cm under the load of the block. The block is given a sharp impulse from below so that it acquires an upward speed 2.0 m/s. How high will it rise? Take .
Let be the spring constant of the spring. In the initial position the forces are the ) the spring tension. But the net force in the initial gravitational force and – ( position is zero. ∴ ⇒ Now suppose the spring rises to a height above the initial position. At this height the velocity and thefore kinetic energy is zero. If we take gravitational potential energy at
8
Work and Energy the initial position as zero, the same at the final position will be . Since no external force does any work, by the law of conservation of mechanical energy we have . ( ) i.e. ( ) ⇒ ⇒ ⇒ ⇒
42. A block of mass 250 g is kept on a vertical spring of spring constant 100 N/M fixed from below. The spring is now compressed to have a length 10 cm shorter than its natural length and the system is released from this position. How high does the block rise? Take .
Let be the height that the block gains above its initial position and be the compression of the spring at the initial position. Since no external force does any work on the block–spring–earth system, by the principle of conservation of mechanical energy we have …I where is the total gravitational and elastic potential energy. Let the gravitational potential energy of the block at the initial position be zero. Taking as the configuration, when spring compression is and as the configuration, when spring compression is zero, we get … II where is the velocity of the block at the position when spring is unstretched. Now we consider block-earth (without spring) system. At the height the block has already separated from the spring and therefore spring does not exert any force on it. At this point velocity of the block is zero and therefore its kinetic energy is also zero. Now we take as the system configuration when spring is unstretched and as the configuration when block is at the maximum height . By law of conservation of mechanical energy we have … III From II and III, we now get ( (
) )(
)
43. Figure shows a spring fixed at the bottom end of an incline of inclination 37°. A small block of mass 2 kg starts slipping down the incline from a point 4.8 m away from the spring. The block compresses the spring by 20 cm, stops momentarily and then rebounds through a distance of 1 m up the incline, find (a) the friction coefficient between the plane and the block and (b) the spring constant of the spring. Take .
(a) Let μ be the friction coefficient and be the force of friction. Then cos Let the reference point for the gravitational potential energy to be zero be at the
8
Work and Energy lowest point. When the block starts motion 4.8 m above spring i.e. 5.0 m above the lowest point along the incline its only energy is the gravitational potential energy ( sin ) At the lowest point the gravitational potential energy being zero, the only energy the block-spring-plane system has is the elastic potential energy of the spring. The work
done by the frictional force
during the descent of the block is
But ⇒ …I Again as the block rises 1 m above the lowest point along the incline, the work done by the friction is The gravitational potential energy, which is the only energy associated with the system is (sin ) But K U ⇒ … II Solving I and II for μ we have (b) From =n I of part (a) we have 44. A block of mass m moving at a speed compresses a spring through a distance speed is halved. Find the spring constant of the spring.
before its
The elastic potential energy of the spring is zero at the initial position. Therefore the kinetic energy at the initial point and the sum of kinetic energy and elastic potential energy after the compression must be equal ∴
. /
⇒ 45. Consider the situation shown in figure. Initially the spring is unstretched when the system is released from rest. Assuming no friction in the pulley find the maximum elongation of the spring.
Let the reference point for the gravitational potential energy to be zero be the initial position of the block. After elongation of the spring. By the principle of conservation of mechanical energy we have ⇒
8
Work and Energy
46.
k1
m
k2
A block of mass m is attached to two unstretched springs of spring constants as shown in figure. The block is displaced towards right through a distance and is released. Find the speed of block as it passes through the mean position shown.
When the block is displaced to the right through a distance , the total energy of the system at rest is ( ) When the block passes through the mean position, the total energy of the block spring system is By the principle of conservation of mechanical energy, we get ( ) ⇒
√ ⃗⃗ v
47.
k
m
A block of mass m sliding on a smooth horizontal surface with a velocity ⃗ meets a long horizontal spring fixed at one end and having spring constant as shown in figure. Find the maximum compression of the spring. Will the velocity of the block be the same as ⃗ when it comes back to the original position shown.
When the block meets the unstretched spring with velocity v ⃗⃗, the total energy of the spring-block system is …I Next suppose the maximum compression of the spring is and at this compression the velocity of the block is zero and therefore the total energy of the system is … II By the principle of conservation of mechanical energy read with =ns I and II we have √ No, the velocity at the original position will be equal in magnitude but opposite in direction.
8
Work and Energy
48.
d A small block of mass 100 g is pressed against a horizontal spring fixed at one end to compress the spring through 5.0 cm. The spring constant is 100 N/m. When released, the block moves horizontally till it leaves the spring. Where will it hit the ground 2 m below the spring?
Suppose is the velocity when the block leaves the spring and starts a projectile motion projected with velocity at an angle 0° with the horizontal. By the principle of conservation of mechanical energy we get Putting in values of √
(
and and solving for )
Next considering the horizontal component of the projectile motion and noting that the initial vertical velocity is zero and acceleration we get ⇒
√ =0.63 s
Since the block is projected with a horizontal velocity of 1.58 m/s and acceleration is zero we get 49. A small heavy block is attached to the lower end of a light rod of length which can be rotated about its clamped upper end. What minimum horizontal velocity should the block be given so that it moves in a complete vertical circle?
If the block moves in a completes vertical circle, then By the principle of conservation of mechanical energy, taking the lowest and the uppermost points as the initial and final positions of the block, we get ( ) ⇒
√
50. Figure shows two blocks A and B each having a mass of 320 g connected by a light string passing over a smooth light pulley. The horizontal surface on which the block A can slide is smooth. The block A is attached to a spring of spring constant 40 N/m whose other end is fixed to a spot 40 cm above the horizontal surface. Initially the spring is vertical and unstretched when the system is released to move. Find the velocity of the block A at the instant it breaks off the surface below it. Take .
Suppose the block A displaces through a distance before it breaks away from the surface. Let the elongation of the spring be . At this break away point the total energy of the two blocks-spring system is
8
Work and Energy (
) ( ) where is the depth of the block B at the initial moment. By the principle of conservation of mechanical energy this energy will be equal to total energy of the system at the initial point which is because there is no compression or elongation of the spring and there is no kinetic energy either. ⇒ …I Next suppose the spring makes an angle θ with the vertical at the break away point Then by taking vertical component of the spring force and the gravitation force on the block A we have ( ) cos ( ) ⇒ (
⇒ But
)
… II )
√(
(
)
Now from I √
(
)(
)
(
)
51. One end of a spring of natural length and spring constant is fixed at the ground and the other is fitted with a smooth ring of mass which is allowed to slide on a horizontal rod fixed at a height . Initially, the spring makes an angle of 37° with vertical when the system is released from rest. Find the speed of the ring when the spring becomes vertical.
Let be the initial elongation of the spring. Then cos ⇒ The mechanical energy at the initial position is And the mechanical energy at the position when spring is vertical is By the principle of conservation of mechanical energy ⇒
√
which
√
52. Figure shows a light rod of length rigidly attached to a small heavy block at one end and a hook at the other end. The system is released from rest with the rod in a horizontal position. There is a fixed smooth ring at a depth below the initial position of the hook and the hook get into the ring as it reaches there. What should be the minimum value of so that the block moves in a complete circle about the ring ?
As the block-rod-hook system free-falls for depth , the velocity point is
of the block at that
8
Work and Energy √ Taking this configuration as the initial position with zero gravitational potential energy of the block and the configuration when the block is above this position of the block as the final position, we have By the principle of conservation of mechanical energy (√
)
⇒ 53. The bob of a pendulum at rest is given a sharp hit to impart a horizontal velocity √ where is the length of the pendulum. Find the tension in the string when (a) the string is horizontal, (b) the bob is at its highest point and (c) string makes an angle of 60° with the upward vertical.
(a) Let the initial configuration be the reference configuration with zero potential energy. Then just after the velocity has been imparted to the bob, the mechanical energy of the bob-string system is Let be the velocity when the string be horizontal. Then By the principle of conservation of mechanical energy ⇒ √ Then let T be the tension in the string when it is horizontal. At this point there is no vertical force and thus (b) At the highest point, let the velocity be
and the mechanical energy is given by
By the principle of conservation of mechanical energy √ (c)
⇒ When the pendulum makes an angle 60° with the upward vertical, height is cos If is the velocity at this point, then ( ) ⇒ ⇒
√ cos
54. A simple pendulum consists of a 50 cm long string connected to a 100 g ball. The ball is pulled aside so that the string makes an angle of 37° with the vertical and is then released. Find the tension in the string when the bob is at its lowest position.
8
Work and Energy Let represent the configuration when the string makes an angle 37° with the vertical and the configuration when the string is at the lowest position. Then, if is the initial height of the bob above lowest point, By the principle of conservation of mechanical energy * ( )+ ⇒ cos The tension at the lowest point consists of the centrifugal force and the gravitational force. ∴
. /
55. Figure shows a smooth track, a part of which is a circle of radius R. A block of mass is pushed against a spring of spring constant fixed at the left end and is trhen released. Find the initial compression of the spring so that the block presses the track with force when it reaches the point P, where the radius of the track is horizontal.
Let be the initial compression of the spring and let be the initial configuration at this compression. Let be the final configuration when block is at point P. When the block is at point P the radius is parallel to the horizontal and therefore the centrifugal force on the block is ⇒ ∴ Then By the principle of conservation of mechanical energy ⇒
√
56. The bob of a stationary pendulum is given a sharp hit to impart it a horizontal speed of √ find the angle rotated by the string before it becomes slack.
,
8
Work and Energy
𝜃
𝜋
𝜋
𝜃
𝜃 𝑙
The string will become slack when the string makes an angle θ with the downward vertical and vertical component of the centrifugal force is equal to mg. Let be the initial configuration immediately after the sharp hit and the be the final configuration when string makes angle θ with the downward vertical Then cos( cos
⇒
)
.
sin .
cos
(
// cos )
. cos / By the principle of conservation of mechanical energy ⇒ . cos / ⇒ cos ⇒ cos . / 57. A heavy particle is suspended by a 1.5 m long string. It is given a horizontal velocity of √ m/s. (a) Find the angle made by the string with the upward vertical, when it becomes slack. (b) Find the speed of the particle at this instant. (c) Find the maximum height reached by the particle over the point of suspension. Take .
(a) Let and represent initial i.e. when particle is at starting point and final configurations when the string becomes slack. Suppose it makes an angle θ with the upward vertical at the final position. Then cos ⇒
cos
8
Work and Energy ( cos ) By the principle of conservation of mechanical energy cos ( cos ) ⇒ cos ⇒ cos (b) From part (a) √( (c)
)(
). /
When the string becomes slack, the particle starts projectile motion. If is the maximum height above the point of suspension reached by the particle then cos
(
)
( ) (
)
58. A simple pendulum of length L having a bob of mass is deflected from its rest position by an angle θ and released The string hits a peg which is fixed at a distance below the point of suspension and the bob starts going in a circle centered at the peg. (a) Assuming that initially the bob has a height less than the peg. Show that the maximum height reached by the bob equals its initial height (b) If the pendulum is released with θ and find the maximum height reached by the bob above its lowest position before the string becomes slack. (c) Find the minimum value of for which the bob goes in a complete circle about the peg when the pendulum is released from θ
(a) θ
x
L
L(1- cos θ)
Let the lowest position reached by the bob be the reference position where gravitational potential energy of the bob-earth system is zero. Take the configuration, with the bob at the initial position, as and the highest position, above the reference position, reached by the bob after the string hits the peg as the final configuration . Then, since , we have ( cos ) and By the principle of conservation of mechanical energy, we have ( cos ) ( ⇒ cos ) [Q. E. D.] (b) In this case let be the configuration, with the bob at a height above the reference position, when string becomes slack. Let be the speed of the bob at the final position. Then
8
Work and Energy
⇒ ( ) …I Now let ϕ be the angle that the string makes with the horizontal at the point . Then equating the centrifugal force and the component of the gravitational force along the string, we get sin
… II
But sin … III and from I, II and III
Eliminating
( (c)
)
⇒ In this case the pendulum is released from the initial position where . The peg is a distance below the point of suspension of the pendulum. Let be the velocity of the bob at the highest point taken as the final position . The lowest point being below the peg, the highest point, the bob reaches, is ( ) above the lowest point. By the principle of conservation of mechanical energy ( ) Also at the highest point, the centrifugal force equals the gravitational force. ⇒ ⇒
(
)
(
)
⇒ 59. A particle slides on the surface of a fixed smooth sphere starting from the top most point. Find the angle rotated by the radius through the particle when it leaves contact with the sphere.
θ
While the particle goes round the sphere, it is subject to a centripetal force and the gravitational force. At the point it loses contact with the sphere, the component of the gravitational force along the radius is equal to the centripetal force. Thus cos
8
Work and Energy Also by the principle of conservation of mechanical energy ( cos ) ⇒
cos
⇒
cos
60. A particle of mass is kept on a fixed smooth sphere of radius R at a position when the radius through the particle makes an angle of 30° with the vertical. The particle is released from this position (a) what is the force exerted by the sphere on the particle just after the release? (b) Find the distance travelled by the particle before it leaves contact with the sphere.
(a) The force exerted by the sphere on the particle is √
cos (b)
θ
Let the lowest point of the sphere be the reference point, with zero potential gravitational energy. Let us assume that the particle revolves an angle θ before it loses contact with the sphere. Then ( ) * )+ cos cos( At the point it loses contact with the sphere cos( ⇒ ⇒
)
cos( cos
) √
√
61. A particle of mass is kept on the top of a smooth sphere of radius R. It is given a sharp impulse which imparts it a horizontal speed . (a) Find the normal force between the sphere and the particle just after the impulse. (b) What should be the minimum value of for which the particle does not slip on the sphere? (c) Assuming the velocity to be half the minimum calculated in part (b) find the angle made by the radius through the particle with the vertical when it leaves the sphere.
(a) Just after the impulse, the horizontal velocity of the particle is and the centripetal force on the particle is . Since both the centripetal force and the gravitational force act in the vertical direction, the normal force of the sphere on the particle is
8
Work and Energy (b) The minimum value of for which the particle does not slip on the sphere is such that normal force is zero. Thus from part (a) ⇒ (c)
√ √
In this case
⇒ Let θ be the angle revolved by the radius when the particle leaves the sphere. Let be the speed of the particle at this point. Then cos By the principle of conservation of mechanical energy ( cos ) ⇒ ⇒ ⇒
cos
(
cos )
cos cos
62. Figure shows a smooth track which consists of straight inclined part of length joining smoothly with the circular part. A particle of mass is projected up the incline from its bottom. (a) Find the minimum projection speed for which the particle reaches the top of the track. (b) Assuming that the projection speed is and that the block does not lose contact with the track before reaching its top, find the force acting on it when it reaches the top. (c) Assuming that the projection speed is only slightly greater than , where will the block lose contact with the track?
(a)
θ
θ
Let be the velocity of the block when it reaches the top of the circular part. Let R be the radius of the circle of the circular part and be the height of the topmost point of the track from the initial starting point. Then sin ( cos ) … I If the particle just reaches the topmost point of the track, its speed thereat will be zero. Therefore by the principle of conservation of mechanical energy, we get … II ⇒ cos ) sin √ , ( (b) In this case the project speed is . Let the speed of block when it reaches the top of the track be . Therefore using the =n I of part (a), we have by the principle of conservation of mechanical energy
8
Work and Energy (
)
But the force F at the top of the track is the centripetal force part (a) we get
. Using
II of
⇒ (c)
⇒ . cos sin / If the initial speed is slightly greater than then the block will move past the highest point of the track . Now suppose the radius through the block makes an angle ϕ with the vertical when the block loses contact with the track. cos Then by the principle of conservation of mechanical energy , sin ( cos ) ( cos ), sin ( cos )cos sin ( cos ) ( cos ( ⇒ cos ) cos ) ⇒ cos
63. A chain of length and mass lies on the surface of a smooth sphere of radius R > with one end tied to the top of the sphere. (a) Find the gravitational potential energy of the chain with reference level at the center of the sphere. (b) Suppose the chain is released and slides down the sphere find the kinetic energy of the chain, when it has slid through an angle θ (c) Find the tangential acceleration of the chain when the chain starts sliding down.
(a) Let be an element of the chain length. The mass of the element is . Let the element make an angle at the center of the sphere Let θ be the angle which the radius passing through the element makes with the vertical. Let be the potential energy of the chain element. Then cos But ⇒
∫
cos
,sin sin (b) Since the chain has slid through an angle θ the final potential energy of the chain is given by ∫
cos
.sin . / sin / By the principle of conservation of mechanical energy 0sin
sin
sin .
/1
8
Work and Energy (c)
From part (b) we have ,sin
sin
sin(
0cos
cos .
)-
Differentiating But
/1
and when the chain starts moving θ 0
∴
1
.
cos /
64. A smooth sphere of radius R is made to translate in a straight line with a constant acceleration . A particle kept on the top of the sphere is released from there at zero velocity with respect to the sphere. Find the speed of the particle with respect to the sphere as a function of the angle θ it slides A P
B θ
m
v
O
Let frame of reference be the one attached to the sphere. With zero velocity of the particle relative to sphere at point A and center of the sphere as the reference point for potential energy, the total mechanical energy of the system at A is The total mechanical energy at the final position (point P) where the particle makes an angle θ with the vertical is cos Since the reference frame attached to the sphere accelerates, there is a pseudo force opposite to the direction of . The work done by this force is sin By the principle of conservation of mechanical energy ⇒
sin
, ( sin ⇒ This is an alternative solution
(
cos )
cos )-
Let be the velocity of the particle relative to the sphere when it has slid through an angle θ Then taking initial velocity of the sphere to be zero the horizontal component of the velocity of the particle is
8
Work and Energy …I √ Since the sphere is smooth, there will not be any force of friction and the downward acceleration will be . The initial vertical velocity is zero and the distance covered is ( cos ). ( ∴ cos ) … II √ From I and II above √
,
* sin
(
cos )+∎
