Experts define power as when work is completed or the transferring of energy happens. They define a force’s average power as the ratio of the work, W, to the total time t taken.

P_{aw} = W/t

One can define instantaneous power as the limiting value of the average power as the approaching of time interval takes place to zero,

P = dW/dt

Here, the work dW takes place by a force F, thereby leading to a displacement dr is dW = F.dr. The expression of the instantaneous power can also take place as

P = F.dr/dt.

Here, v represents the instantaneous velocity when F represents the force. The **Work Energy And Power Class 11 NCERT Solutions** will help you better understand this concept.

**NCERT Solutions For Class 11 Physics Chapter 6 Work, Energy and Power**

Below are some **NCERT Solutions For Class 11 Physics Chapter 6:**

**1. The movement body of mass 2 kg, initially at rest, occurs due to an applied horizontal force of 7N. This movement takes place on a table whose coefficient of kinetic friction is 0.1.Now, calculate the following:**

**(a) work completed due to the applied force in 10 seconds,**

**(b) work done due to friction in 10 seconds,**

**(c) work done due to the net force acting on the body in 10 seconds,**

**(d) change that takes place in the kinetic energy of the body in 10 s,**

**Ans: **

The body’s mass = 2 kilograms

Applied horizontal force = 7 Newton

Kinetic friction’s coefficient = 0.1

Acceleration whose production take place by the applied force, a1= F/m = 7/2 = 3.5 m/s^{2}

Force of friction in this solution would be, f = μR = μmg = 0.1 \( \times \) 2 \( \times \) 9.8

Retardation whose production takes place by friction, a_{2} = -f/m = -196/2 = -0.98

Net acceleration with which the movement of the body takes place

a = a_{1} + a_{2} = 3.5 – 9.8 = 2.5

Distance that the body moves in 10 seconds,

s = ut + (1/2)at2 = 0 + (1/2) \( \times \) 2.52 \( \times \) (10)2= 126 m

(a) The time at which the determination of the work has to take place is t = 10 s.

The formula of Work will be = Force \( \times \) displacement

= 7 \( \times \) 126 = 882 J

(b) Work that takes place by the friction in 10 seconds is represented by w, which is

-f \( \times \) s = -1.96 x 126

(c) Work that takes place by the net force in 10 seconds is represented by w, which is

(F – f)s, which turns out to be = (7 – 1.96) 126 = 635 J

(d) From v = u + at

V, whose value turns out to be = 0 + 2.52 \( \times \) 10 = 25.2 m/s

So, the Final Kinetic Energy in this solution would be = (1/2) mv2 = (1/2) \( \times \) 2 \( \times \) (25.2)2 = 635 J

Initial Kinetic Energy in this NCERT solution would be = (1/2) mu2 = 0

The change that takes place in the Kinetic energy = 635 – 0 = 635 J

The work that takes place by the net force shall be equal to the final kinetic energy.

**2. State if each of the following statements in the following NCERT solution is true or false. Provide reasons for whatever answer you give.**

**a. If an elastic collision takes place between two bodies, the momentum and energy of each body shall be considered to be conserved.**

**Ans:** False

The Law of conservation of momentum clearly states that a system’s total momentum is conserved when no application of net external force takes place on the system. The transferring of the momentum of one body can take place to the other body in the system.

**b. The total energy of a system shall always be conserved, irrespective of the application of the internal and external forces on the body.**

**Ans:** False

Any external force on a body can bring about a change in the body’s total energy.

**c. Work that takes place in the motion of a body over a closed-loop shall be zero for every possible natural force.**

**Ans:** False

Total energy over a closed-loop motion shall be zero only if conservative force is involved.

**d. In an inelastic collision in a system, the final kinetic energy shall always be less than the initial kinetic energy.**

**Ans:** True

In inelastic collisions, the total energy is usually lost in the form of vibrations of sound or heat. Consequently, the final kinetic energy shall be less than the initial kinetic energy.

**3. With a speed of 200m/s, a molecule in a gas container hits a horizontal wall at an angle 30 degrees with the normal. Afterwards, its rebounding takes place with the same speed. Does conservation of momentum take place in the collision? Also, tell whether the collision would be elastic or inelastic?**

**Ans:** The conservation of a system’s momentum shall occur as long as no external forces are acting on the system. Similarly, the conservation of the system’s momentum shall happen in the gas container. Here, it does not matter whether there has been an elastic collision or inelastic collision.

As the molecule strikes against the container’s wall, its rebounding takes place at the same speed. Therefore, the wall’s velocity shall continue to be zero. Therefore, the collision over here would be elastic. Consequently, the molecule’s total kinetic energy during the collision remains conserved.