Welcome to the Chapter 6 - Work Energy and Power, Class 11 Physics NCERT Solutions page. Here, we provide detailed question answers for Chapter 6 - Work Energy and Power. The page is designed to help students gain a thorough understanding of the concepts related to natural resources, their classification, and sustainable development.
Our solutions explain each answer in a simple and comprehensive way, making it easier for students to grasp key topics Work Energy and Power and excel in their exams. By going through these Work Energy and Power question answers, you can strengthen your foundation and improve your performance in Class 11 Physics. Whether you’re revising or preparing for tests, this chapter-wise guide will serve as an invaluable resource.
(a) Positive In the given case, force and displacement are in the same direction. Hence, the sign of work done is positive. In this case, the work is done on the bucket.
(b) Negative In the given case, the direction of force (vertically downward) and displacement (vertically upward) are opposite to each other. Hence, the sign of work done is negative.
(c) Negative Since the direction of frictional force is opposite to the direction of motion, the work done by frictional force is negative in this case.
(d) Positive Here the body is moving on a rough horizontal plane. Frictional force opposes the motion of the body. Therefore, in order to maintain a uniform velocity, a uniform force must be applied to the body. Since the applied force acts in the direction of motion of the body, the work done is positive.
(e) Negative The resistive force of air acts in the direction opposite to the direction of motion of the pendulum. Hence, the work done is negative in this case.
Yes;
Collision is elastic The momentum of the gas molecule remains conserved whether the collision is elastic or inelastic.
The gas molecule moves with a velocity of 200 m/s and strikes the stationary wall of the container, rebounding with the same speed.
It shows that the rebound velocity of the wall remains zero. Hence, the total kinetic energy of the molecule remains conserved during the collision. The given collision is an example of an elastic collision.
Volume of the tank, V= 30 m3
Time of operation, t = 15 min = 15 × 60 = 900 s
Height of the tank, h= 40 m
Efficiency of the pump, n = 30%
Density of water, p = 103 kg/m3
Mass of water, m= pV= 30 ×103 kg
Output power can be obtained as:
P0 = work done / Time = mgh / t
= 30 x 103 x 9.8 x 40 / 900 = 13.067 x 103 W
For input power Pi,, efficiency n is given by the relation:
n = P0 / Pi = 30%
Pi = 13.067 / 30 x 100 x 103
= 43.6 kW
The sand bag is placed on a trolley that is moving with a uniform speed of 27 km/h. The external forces acting on the system of the sandbag and the trolley is zero. When the sand starts leaking from the bag, there will be no change in the velocity of the trolley. This is because the leaking action does not produce any external force on the system. This is in accordance with Newton's first law of motion. Hence, the speed of the trolley will remain 27 km/h.
Mass of the body, m= 2 kg
Applied force, F = 7 N
Coefficient of kinetic friction, µ= 0.1
Initial velocity, u= 0
Time, t = 10 s
The acceleration produced in the body by the applied force is given by Newton's second law of motion as:
a' = F / m = 7/2 = 3.5 m/s2
Frictional force is given as:
f = µmg
= 0.1 × 2 × 9.8 = -1.96 N
The acceleration produced by the frictional force:
a'' = - 1.96 / 2 = -0.98 m/s2
Total acceleration of the body:
a = a' + a''
= 3.5 + (-0.98) = 2.52 m/s2
The distance travelled by the body is given by the equation of motion:
s = ut + ½ at2
= 0 + ½ x 2.52 x 102
= 126 m
(a) Work done by the applied force, Wa= F × s = 7 ×126 = 882 J
(b) Work done by the frictional force, Wf = F× s= -1.96 ×126 = -247 J
(c) Net force = 7 + (-1.96) = 5.04 N Work done by the net force, Wnet= 5.04 ×126 = 635 J
(d) From the first equation of motion, final velocity can be calculated as:
v = u + at = 0 + (25.2)2 ×10 = 25.2 m/s
Change in kinetic energy = 1/2 mv2 - 1/2 mu2
= ½ x 2(v2 - u2) = (25.2)2 - 02 = 635 j
Area of the circle swept by the windmill = A
Velocity of the wind = v
Density of air = p
(a) Volume of the wind flowing through the windmill per sec = Av
Mass of the wind flowing through the windmill per sec = pAv
Mass m, of the wind flowing through the windmill in time t = pAvt
(b) Kinetic energy of air = ½ mv2
= ½ (pAvt)v2 = ½ pAv3t
(c) Area of the circle swept by the windmill = A = 30 m2
Velocity of the wind = v= 36 km/h
Density of air, p = 1.2 kg m-3
Electric energy produced = 25% of the wind energy
= 25/100 x Kinetic energy of air
= 1/8 pAv3t
Electrical Power = Electical energy / time
= 1/8 pAv3t / t = 1/8 pAv3
= 1/8 x 1.2 x 30 x 103
= 4.5 x 103 w
= 4.5 kW
(a) Mass of the weight, m = 10 kg
Height to which the person lifts the weight, h = 0.5 m
Number of times the weight is lifted, n = 1000
∴Work done against gravitational force:
= n(mgh)
= 1000 x 10 x 9.8 x 0.5
= 49 x 103J = 49kJ
(b) Energy equivalent of 1 kg of fat = 3.8 × 107 J
Efficiency rate = 20%
Mechanical energy supplied by the person's body:
= 20/100 x 3.8 x 107 J
= 1/5 x 3.8 x 107 J
Equivalent mass of fat lost by the dieter:
= 1 / (1/5 x 3.8 x 107) x 49 x 103
= 245 / 3.8 x 10-4
= 6.45 x 10-3 kg
(a) 200 m2
(a) Power used by the family, P = 8 kW = 8 × 103 W
Solar energy received per square metre = 200 W
Efficiency of conversion from solar to electricity energy = 20 %
Area required to generate the desired electricity = A
As per the information given in the question, we have:
8 x 103 = 20% x (A × 200)
= 20/100 x A x 200
∴ A = 8 x 103 / 40 = 200 m2
(b) The area of a solar plate required to generate 8 kW of electricity is almost equivalent to the area of the roof of a building having dimensions 14 m × 14 m.
Mass of the bullet, m= 0.012 kg
Initial speed of the bullet, ub= 70 m/s
Mass of the wooden block, M= 0.4 kg
Initial speed of the wooden block, uB= 0
Final speed of the system of the bullet and the block = v
Applying the law of conservation of momentum:
mub + muB = (m + M) v
0.012 × 70 + 0.4 × 0 = (0.012 + 0.4)v
∴ v = 0.84 / 0.412 = 2.04 m/s
For the system of the bullet and the wooden block:
Mass of the system, m' = 0.412 kg
Velocity of the system = 2.04 m/s
Height up to which the system rises = h
Applying the law of conservation of energy to this system:
Potential energy at the highest point = Kinetic energy at the lowest point
m'gh = ½ m'v2
= ½ (2.04)2 / 9.8
= 0.2123 m
The wooden block will rise to a height of 0.2123 m.
Heat produced = Kinetic energy of the bullet - Kinetic energy of the system
½ mu2 - ½ m'v2
= 29.4 - 0.857 = 28.54 J
Mass of the bolt, m= 0.3 kg
Speed of the elevator = 7 m/s
Height, h = 3 m
Since the relative velocity of the bolt with respect to the lift is zero, at the time of impact, potential energy gets converted into heat energy.
Heat produced = Loss of potential energy
= mgh = 0.3 x 9.8 x 3
= 8.82 J
The heat produced will remain the same even if the lift is stationary. This is because of the fact that the relative velocity of the bolt with respect to the lift will remain zero.
(a) Decreases
(b) Kinetic energy
(c) External force
(d) Total linear momentum
Explanation: (a) A conservative force does a positive work on a body when it displaces the body in the direction of force. As a result, the body advances toward the centre of force. It decreases the separation between the two, thereby decreasing the potential energy of the body.
(b) The work done against the direction of friction reduces the velocity of a body. Hence, there is a loss of kinetic energy of the body.
(c) Internal forces, irrespective of their direction, cannot produce any change in the total momentum of a body. Hence, the total momentum of a many- particle system is proportional to the external forces acting on the system.
(d) The total linear momentum always remains conserved whether it is an elastic collision or an inelastic collision.
(a) False (b) False (c) False (d) True
Explanation: (a) In an elastic collision, the total energy and momentum of both the bodies, and not of each individual body, is conserved.
(b) Although internal forces are balanced, they cause no work to be done on a body. It is the external forces that have the ability to do work. Hence, external forces are able to change the energy of a system.
(c) The work done in the motion of a body over a closed loop is zero for a conservation force only.
(d) In an inelastic collision, the final kinetic energy is always less than the initial kinetic energy of the system. This is because in such collisions, there is always a loss of energy in the form of heat, sound, etc.
(a) No
In an elastic collision, the total initial kinetic energy of the balls will be equal to the total final kinetic energy of the balls. This kinetic energy is not conserved at the instant the two balls are in contact with each other. In fact, at the time of collision, the kinetic energy of the balls will get converted into potential energy.
(b) Yes
In an elastic collision, the total linear momentum of the system always remains conserved.
(c) No; Yes
In an inelastic collision, there is always a loss of kinetic energy, i.e., the total kinetic energy of the billiard balls before collision will always be greater than that after collision.
The total linear momentum of the system of billiards balls will remain conserved even in the case of an inelastic collision.
(d) Elastic
In the given case, the forces involved are conservation. This is because they depend on the separation between the centres of the billiard balls. Hence, the collision is elastic.
(ii) t
Mass of the body = m
Acceleration of the body = a
Using Newton's second law of motion, the force experienced by the body is given by the equation:
F = ma
Both m and a are constants.Hence, force F will also be a constant.
F = ma = Constant … (i)
For velocity v, acceleration is given as,
a = dv / dt = constant
dv = Constant x dt
v = αt ....... (ii)
where α is another constant
Power is given by the relation:
P = F.v
Using equations (i) and (iii), we have:
P ∝ t
Hence, power is directly proportional to time.