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NCERT Solutions for Class 11 Physics Physics Chapter 6 Work Energy and power
NCERT Solutions for Class 11 Physics Physics Chapter 6 Work Energy and power are been solved by expert teachers of CBSETuts.com. All the solutions given in this page are solved based on CBSE Syllabus and NCERT guidelines.
The sign of work done by a force on a body is important to understand. State carefully if the following quantities are positive or negative:
(a) work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket.
(b) work done by gravitational force in the above case,
(c) work done by friction on a body sliding down an inclined plane,
(d) work done by an applied force on a body moving on a rough horizontal plane with uniform velocity,
(e) work done by the resistive force of air on a vibrating pendulum in bringing it to rest.
Work done W = F-dcosθ
where θ is the angle between the direction of force vector and displacement vector .
(a) Work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket is positive because lifting a bucket out of a well, force equal to the weight of the bucket has to be applied by the man along the vertical in upward direction. Since displacement is also in upward direction θ= 0°. Therefore
W = F.dcosθ = F.dcos0° = Fd. (positive).
(b) Work done by gravitational force in the above case is negative because the bucket moves in a direction opposite to the gravitational force which is acting vertically downwards. The angle between the gravitational force and the displacement is 180°. Therefore,
W= F.dcosθ = F.dcos180° = – F.d. (negative).
(c) Work done by friction on a body sliding down an inclined plane is negative. Friction always acts in a direction opposite to the direction of motion. Therefore 0 = 180°,
W = F-dcos0 = F.dcos180°
= -Fd. (negative).
(d) Work done by an applied force on a body moving on a rough horizontal plane with uniform velocity is positive. Because applied force and the displacement are in same direction. Therefore
θ= 0°, W= Fdcosθ = Fdcos0°
= Fd (positive).
(e) Work done by the resistive force of air on a vibrating pendulum in bringing it to rest is negative because direction of resistive force is opposite to the direction of motion of the pendulum. Therefore
θ = 180°, W = Fdcosθ = F.dcos180°
= -Fd (negative).
A body of mass 2 kg initially at rest moves under the action of an applied horizontal force of 7 N on a table with coefficient of kinetic friction = 0.1. Compute the
(a) work done by the applied force in 10 s,
(b) work done by friction in 10 s,
(c) work done by the net force on the body 10 s,
(d) change in kinetic energy of the body in 10 s. Interpret your result.
Given, m = 2 kg, u=0,m = 0.1, t = 10 s
Applied force F = 7 N
Force due to friction f = μmg
= 0.1 x 2 x 9.8 = 1.96 N
Net force under which body moves F’ = F – f
= 7-1.96 = 5.04 N
Therefore acceleration with which body moves
Change in kinetic energy = Final K.E. – Initial K.E. = 635 – 0 = 635 This shows that change in kinetic energy of the body is equal to work done by net force on the body.
Given figures are examples of some potential energy functions in one dimension. The total energy of the particle is indicated by a cross on the ordinate axis. In each case, specify the regions, if any, in which the particle cannot be found for the given energy. Also, indicate the minimum total energy the particle must have in each case. Think of simple physical contexts for which these potential energy shapes are relevant.
Total energy = Kinetic energy + Potential energy
Kinetic energy can never be negative. The object can exist in the region, which is K.E. would become positive.
- In the region between x = 0 to x = a,E. is zero. Therefore kinetic energy in this region is positive. However, in the region x > a, the potential energy has a value V0 > E, therefore kinetic energy becomes negative. Hence the object cannot exist in this region x > a.
- The object cannot be present in any region because potential energy (V0) > E in every region.
- In this regions between x = 0 to x = a and x > b, the potential energy (V0) is greater than total energy E of the object. Therefore kinetic energy becomes negative the object cannot be present in the x < a and x>b.
- The object cannot exist in the region -b/2 <x<-a/2 and a/2 <x< b/2.
The potential energy function for a particle executing linear simple harmonic motion is given by
V(x) = kx2/2, where k is the force constant of the oscillator. For k = 0.5 Nm1, the graph of V(x) versus x is shown in figure. Show that a particle of total energy 1 J moving under this potential must turn back when it reaches x = ± 2 m.
The total energy of an oscillator is the sum of kinetic energy and potential energy at any instant.
The particle turn back at the instant, when its velocity becomes zero, u = 0,
Thus, the particle of total energy 1 J moving under this potential, must turn back at
x = ± 2 m.
Answer the following :
(a) The casing of a rocket in flight burns up due to friction. At whose expense is the heat energy required for burning obtained? The rocket or the atmosphere?
(b) Comets move around the sun in highly elliptical orbits. The gravitational force on the comet due to the sun is not normal to the comet’s velocity in general. Yet the work done by the gravitational force over every complete orbit of the comet is zero. Why?
(c) An artificial satellite orbiting the earth in very thin atmosphere loses its energy gradually due to dissipation against atmospheric resistance, however small. Why then does its speed increase progressively as it comes closer and closer to the earth?
(d) In figure
(1) the man walks 2 m carrying mass of 15 kg on his hands. In figure
(2) he walks the same distance pulling the rope behind him. The rope goes over a pulley, and a mass of 15 kg hangs at its other end. In which case is the work done greater?
(a) When the casing burn up, its mass decreases due to this total energy of the rocket decreases because total energy of rocket in flight depends on its mass. Hence heat energy required for burning is obtained from the rocket itself and not from the atmosphere.
(b) This is because gravitational force is conservative force. Work done by the gravitational force of the sun over a closed path in every complete orbit of the comet is zero.
(c) When the artificial satellite orbiting the earth comes closer and closer to earth, its potential energy decreases. As sum of potential energy and kinetic energy is constant, therefore, K.E. of satellite and hence its velocity goes on increasing. However, total energy of the satellite decreases a little on account of dissipation against atmospheric resistance.
(d) In figure (i), force is applied on the mass, by the man in vertically upward direction but distance is moved along the horizontal
∴ θ = 90°.
W= F s cos 90° = Zero.
In figure (ii), force is applied along the vertical and the distance moved is also along the vertical.
W = F s cosθ = mg x s cos0°
W = 15 x 9.8 x 2 x 1 = 294 joule
∴ Work done is greater in 2nd case.
Underline the correct alternative:
(a) When a conservative force does positive work on a body, the potential energy of the body increases/decreases/remains unaltered.
(b) Work done by a body against friction always results in a loss of its kinetic/ potential energy.
(c) The rate of change of total momentum of a many-particle system is proportional to the external force/sum of the internal forces on the system.
(d) In an inelastic collision of two bodies, the quantities which do not change after the collision are the total kinetic energy/total linear momentum/total energy of the system of two bodies.
(a) Potential energy of the body decreases. The conservative force does positive work on a body, when it displaces the body in the direction of force. The body, therefore, approaches the center of force, decreasing x. Hence, P.E. decreases.
(b) Work is done by a body against friction at the expense of its kinetic energy. Hence K.E. of the body decreases.
(c) Internal forces cannot change the total or net momentum of a system. Hence the rate of change of total momentum of many particle system is proportional to the external force on the system.
(d) In an inelastic collision of two bodies, the quantities which do not change after the collision are total linear momentum and total energy of the system of two bodies. The total K.E. changes as some energy appears in other forms.
State if each of the following statements is true or false. Give reasons for your answer.
(a) In an elastic collision of two bodies, the momentum and energy of each body is conserved.
(b) Total energy of a system is always conserved, no matter what internal and external forces on the body are present.
(c) Work done in the motion of a body over a closed loop is zero for every force in nature.
(d) In an inelastic collision, the final kinetic energy is always less than the initial kinetic energy of the system.
(a) False : The total momentum and total energy of the system are conserved.
(b) False : The external forces on the system may increase or decrease the total energy of the system.
(c) False : The work done during the motion of a body over a closed loop is zero only when the body is moving under the action of a conservative force, such as gravitational or electrostatic forces. It is not zero when the forces are non conservative such as friction and work done by force of friction is not zero over a closed loop.
(d) True : In an inelastic collision, the final kinetic energy is always less than the initial kinetic energy of the system because in this collision some kinetic energy usually changes into some other form of energy such as heat, sound etc.
Answer carefully, with reasons:
(a) In an elastic collision of two billiard balls, is the total kinetic energy conserved during the short time of collision of the balls (i.e. when they are in contact)?
(b) Is the total linear momentum conserved during the short time of an elastic collision of two balls?
(c) What are the answer to (a) and (b) for an inelastic collision?
(d) If the potential energy of two billiard balls depends only on the separation distance between their centers, is the collision elastic or inelastic? (Note, we are talking here of potential energy corresponding to the force during collision, not gravitational potential energy).
(a) During collision (when balls are in contact), the kinetic energy of balls gets converted into potential energy. Kinetic energy before collision and after collision is the same. Thus during the given elastic collision, the total kinetic energy is not conserved.
(b) Yes, the total linear momentum is conserved during the short time of an elastic collision of two balls.
(c) The total kinetic energy is not conserved in an elastic collision, during collision and after collision. The total momentum of the two balls is conserved.
(d) It is elastic collision.
A body is initially at rest. It undergoes one-dimensional motion with constant acceleration. The power delivered to it at time t is proportional to
(c) t 3/2
A body is moving unidirectionally under the influence of a source of constant power. Its displacement in time t is proportional to
(c) t 3/2
A body constrained to move along the z-axis of a coordinate system is subject to a constant force F given by
where are unit vectors along the x, y and z-axis of the system respectively. What is the work done by this force in moving the body a distance of 4 m along the z-axis?
An electron and a proton are detected in a cosmic ray experiment, the first with kinetic energy 10 keV, and the second with 100 keV. Which is faster, the electron or the proton? Obtain the ratio of their speeds, (electron mass = 9.11 x 10-31 kg, proton mass = 1.67 x 10-27 kg, 1 eV = 1.60 x 10-19 J.)
A rain drop of radius 2 mm falls from a height of 500 m above the ground. It falls with decreasing acceleration (due to viscous resistance of the air) until at half its original height, it attains its maximum (terminal) speed, and moves with uniform speed thereafter. What is the work done by the gravitational force on the drop in the first and second half of its journey? What is the work done by the resistive force in the entire journey if its speed on reaching the ground is 10 m s-1?
Radius of raindrop, r = 2 mm = 2 x 10-3 m
Distance covered by drop in each half of the journey
Mass of raindrop = Volume of drop x Density
= πr3ρ(ρ = 10 kg m = density of water) = 3.35 x 10-? kg
Work done by gravitational force during each half
= mgh = 3.35 x 105 x 9.8 x 250 = 0.082 J
Whether the rain drop falls with decreasing acceleration or with uniform speed, the work done by the gravitational force on the drop remains same.
If there were no resistive force, energy of drop on reaching the ground
E1 = mgh = 3.35 x 10-5 x 9.8 x 500 = 0.164 J
Actual energy, E2 = mv2
= x 3.35 x 10-5 x (10)2 = 1.675 x 10-3 J
Work done by the resistive force W = E2 – E1 = 1.675 x 10-3 -0.164 =-0.162 J
A molecule in a gas container hits a horizontal wall with speed 200 ms-1 and angle 30° with the normal, and rebounds with the speed. Is momentum conserved in the collision? Is the collision elastic or inelastic?
Yes, momentum of the molecule + wall system is conserved. The wall has a recoil momentum such that the momentum of the wall + momentum of the outgoing molecule equals momentum of the incoming molecule, assuming the wall to be stationary initially. However, the recoil momentum produces negligible velocity because of the large mass of the wall. Since kinetic energy is also conserved ,the collision is elastic.
A pump on the ground floor of a building can pump up water to fill a tank of volume 30 m3 in 15 min. If the tank is 40 m above the ground, and the efficiency of the pump is 30%, how much electric power is consumed by the pump.
As given that
Volume of water = 30 m3
Time taken by pump to fill tank = 15 min
= 15 x 60 = 900 s
The height of tank = 40 m
The efficiency of pump = 30%
Consumption of power by pump = ?
Mass of water pumped = Volume x density
= 30 x 103 (Density of water = 103 kg/m3)
Two identical ball bearings in contact with each other and resting on a frictionless table are hit head-on by another ball bearing of the same mass moving initially with a speed v. If the collision is elastic which of the following figure is a possible result after collision?
m is the mass of each ball bearing.
Total kinetic energy of the system before collisions
We observe that kinetic energy is conserved before collision as well as after collision only in Case II. Therefore Case II is the only possibility.
The bob A of a pendulum released from 30° to the vertical hits another bob B of the same mass at rest on a table as shown in figure. How high does the bob A rise after the collision? Neglect the size of the bobs and assume the collision to be elastic.
The bob A transfer its entire momentum to the ball on the table, and the bob A does not rise at all
The bob of a pendulum is released from a horizontal position. If the length of the pendulum is 1.5 m, what is the speed with which the bob arrives at the lowermost point, given that it dissipated 5% of its initial energy against air resistance?
Length of pendulum h= 1.5 m
Let m be mass of the bob, then potential energy of the bob at A,
mgh = m x 9.8 x 1.5 J
On reaching the lowest point B, the bob will acquire an equal amount of kinetic energy. But as 5% of energy is lost against the air friction. Energy converted = 95% (mgh)
If υ is the velocity acquired by the bob at the point B, then
A trolley of mass 300 kg carrying a sandbag of 25 kg is moving uniformly with a speed of 27 km/h on a frictionless track. After a while, sand starts leaking out of a hole on the floor of the trolley at the rate of 0.05 kg s-1. What is the speed of the trolley after the entire sand bag is empty?
The trolley carrying a sand bag is moving with uniform speed of 27 km h Therefore external force on the system (trolley + sand) is zero.
When the sand leaks out of a hole on trolley’s floor, it does not lead to the application of any external force on the trolley. So, the speed of the trolley will not change.
A body of mass 0.5 kg travels in a straight line with velocity v = ax3/2 where a = 5 m_1/2 s_1. What is the work done by the net force during its displacement from x = 0 to x = 2 m?
Here, Mass of body = 0.5 kg
Velocity υ = ax3/2 where a = 5 m_1/2 s_1
Initial velocity at x = 0, υ1 = 5 x 0 = 0
Final velocity at x = 2, υ2 = 5 x 23/2 Work done = Increase in kinetic energy
= x 0.5 [(5×23/2)2-0] J= 50J
The blades of a windmill sweep out a circle of area A.
(a) If the wind flows at a velocity v perpendicular to the circle, what is the mass of the air passing through it in time t?
(b) What is the kinetic energy of the air?
(c) Assume that the windmill converts 25% of the wind’s energy into electrical energy, and that
A = 30 m2, v = 36 km/h and the density of air is 1.2 kg m-3. What is the electrical power produced?
(a) Volume of wind flowing/second = Aυ
Mass of air passing t s = Aυpt
A person trying to lose weight (dieter) lifts a 10 kg mass, one thousand times, to a height of 0.5 m each time. Assume that the potential energy lost each time she lowers the mass is dissipated.
(a) How much work does she do against the gravitational force?
(b) Fat supplies 3.8 x 107 J of energy per kilogram which is converted to mechanical energy with a 20% efficiency rate. How much fat will the dieter use up?
Here, m = 10 kg, h = 0.5 m, n = 1000
(a) Work done against gravitational force
W = n(mgh) = 1000 x (10 x 9.8 x 0.5) = 49000 J
(b) Mechanical energy supplied by 1 kg of fat
A family uses 8 kW of power.
(a) Direct solar energy is incident on the horizontal surface at an average rate of 200 W per square meter. If 20% of this energy can be converted to useful electrical energy, how large an area is needed to supply 8 kW?
(b) Compare this area to that of the roof of a typical house.
Let the area be A square meter.
∴ Total power = 200 A.
Useful electrical energy produced/second
= (200 A) = 40A = 8000 (watt)
Therefore, A = =200 sq. m 40
This area is comparable to the roof of a large house of 250 sq. meter.
A bullet of mass 0.012 kg and horizontal speed 70 ms-1 strikes a block of wood of mass 0.4 kg and instantly comes to rest with respect to the block. The block is suspended from the ceiling by means of thin wires. Calculate the height to which the block rises. Also, estimate the amount of heat produced in the block.
Here, Mass of the bullet m1 = 0.012 kg
Mass of the block m2 = 0.4 kg
Initial velocity of the bullet, μ1= 70 m s-1
initial velocity of the block, μ2 = 0
Since on striking the wooden block, the bullet comes to rest w.r.t. the block of wood, the collision is inelastic in nature.
Let υ be the velocity acquired by the combination.
Applying principle of conservation of linear momentum
Two inclined frictionless tracks, one gradual and the other steep meet at A from where two stones are allowed to slide down from rest, one on each track as shown in figure. Will the stones reach the bottom at the same time? Will they reach there with the same speed? Explain. Given θ1 = 30°, θ2 = 60°, and
h = 10 m, what are the speeds and times taken by the two stones?
Figure shows the two inclined tracks of different lengths but of same heights, such that the angle of inclination θ2 is greater than θ1. The stones along OX and OY will slide down with accelerations
a1 = gsinθ1 and a2 = ysinθ2 respectively. As θ2 > θ1 therefore, a2 > a1.
Now, for motion of the two objects :
K.E. at the bottom = P.E. at the top
or Mgh = Mυ2 or υ= √2gh
As heights of two tracks is same, both the objects will reach the bottom with the same speed.
Now, v= u + at = 0 + at or t=
As υ is same for two objects, t ∝ 1/a. Since a2 > a1 the object sliding on the inclined track OY will reach the bottom earlier. In other words, the two objects will reach the bottom at different times.
A 1 kg block situated on a rough incline is connected to a spring of spring constant 100 N m-1 as shown in figure. The block is released from rest with the spring in the unstretched position. The block moves 10 cm down the incline before coming to rest. Find the coefficient of friction between the block and the incline. Assume that the spring has a negligible mass and the pulley is frictionless.
As is clear from fig
A bolt of mass 0.3 kg falls from the ceiling of an elevator moving down with an uniform speed of 7 m s-1. It hits the floor of the elevator (length of the elevator = 3 m) and does not rebound. What is the heat produced by the impact? Would your answer be different if the elevator were stationary?
Here, m = 0.3 kg, v = 7 m/s, h = length of elevator = 3 m
As relative velocity of the ball w.r.t. elevator is zero, therefore, in the impact, only potential energy of the ball is converted into heat energy.
Amount of heat produced = P.E. lost by the bolt = mgh = 0.3 x 9.8 x 3 = 8.82 J.
The answer shall not be different, if the elevator were stationary as the bolt too in that case would start from stationary position, i.e. relative velocity of the ball w.r.t. elevator would continue to be zero.
A trolley of mass 200 kg moves with a uniform speed of 36 km/h on a frictionless track. A child of mass 20 kg runs on the trolley from one end to the other (10 m away) with a speed of 4 m s-1 relative to the trolley in a direction opposite to the its motion, and jumps out of the trolley. What is the final speed of the trolley? How much has the trolley moved from the time the child begins to run?
Here, mass of trolley, m, = 200 kg
speed of the trolley υ = 36 km/h = 10 m/s
mass of the child, m2 = 20 kg
Before the child starts running, momentum of the system
p1 = (m1 + m2) v = (200 + 20) 10 = 2200 kg m s-1
When the child starts running, with a velocity of 4 m/s in a direction opposite to trolley, suppose υ’ is final speed of the trolley (w.r.t. earth). Obviously, speed of the child relative to earth (υ’ – 4)
Distance moved by the trolley in this time = Velocity of trolley x time = 10.36 x 2.5 = 25.9 m
Which of the following potential energy curves in figure cannot possibly describe the elastic collision of two billiard balls? Here r is the distance between centers of the balls.
The potential energy of a system of two masses varies inversely as the distance (r) between them, i.e., V(r)α . When the two billiard balls touch each other. P.E. become zero i.e. at r = R + R = 2R; V(r) = 0. Out of the given graphs, curve (v) only satisfies these two conditions. Therefore, all other curves cannot possibly describe the elastic collision of two billiard balls.
Consider the decay of a free neutron at rest:
n → p + e–
Show that the two-body decay of this type must necessarily give an electron of fixed energy and, therefore, cannot account for the observed continuous energy distribution to the β-decay of a neutron of a nucleus as shown in the figure.
In the decay process, n —> p + e
energy of electron is equal to (Am)c1
where Δm = mass defect
= mass of neutron – mass of proton and electron;
which is fixed. Therefore, two body decay of this type cannot explain the observed continuous energy distribution in the P-decay of a neutron or a nucleus.
Note : The simple result of this exercise was one among the several arguments advanced by W. Pauli to predict the existence of a third particle in the decay products of β-decay. The particle is known as neutrino. We now know that it is a particle of intrinsic spin 1/2 (like <e–, p or n), but is neutral, and either massless or having an extremely small mass (compared to electron’s mass) and which interacts very weakly with matter. The correct decay process of neutron is n —> p + e + υ’].
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