**NCERT solutions for class 11 physics chapter 6 work energy and power**: Consider that you are carrying a 10 Kg stationary in your hand and you are not moving. In this situation, you are doing work physically according to you. But in physics, you are doing zero work. Can you find why work is zero here? The solutions of NCERT class 11 physics chapter 6 work energy and power** **has questions based on the concepts of work and types of work, the relation between work and energy and the concept of power. In the CBSE NCERT solutions for class 11 physics chapter 6 work energy and power, you will study problems on conservation of energy (energy can be neither created nor be destroyed). NCERT solutions for class 11 physics chapter 6 work energy and power also give answers to the questions based on the concepts of the collision, which are important for the exam. Solutions of NCERT are helpful in scoring well in class and board exams.

6.1 Introduction

6.2 Notions Of Work And Kinetic Energy: The Work-energy Theorem

6.3 Work 6.4 Kinetic Energy

6.5 Work Done By A Variable Force

6.6 The Work-energy Theorem For A Variable Force

6.7 The Concept Of Potential Energy

6.8 The Conservation Of Mechanical Energy

6.9 The Potential Energy Of A Spring

6.10 Various Forms Of Energy: The Law Of Conservation Of Energy

6.11 Power

6.12 Collisions

- For an elastic collision Law of conservation of momentum and that of Kinetic Energy holds good
- For inelastic collision Law of conservation of momentum hold good but kinetic energy is not conserved
- Coefficient of restitution is the ratio of relative velocity after the collision to relative velocity before the collision
- For a perfectly elastic collision coefficient of restitution is one
- For inelastic collision coefficient of restitution is less than one
- For a perfectly inelastic collision coefficient of restitution is zero

**NCERT solutions for class 11 physics chapter 6 work energy and power exercise**

work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket.

In this case, the direction of force and displacement are the same, this work done is **positive**.

In this case, the direction of displacement is upward and the direction of force is downward. Thus work done is **negative** in nature.

We know that friction acts in the direction opposite to the direction of motion. Hence work done by the frictional force is **negative**.

In this case, the applied force supports the motion of the object (balances frictional force). Thus work done by the force is **positive**.

Work done in this case is **negative** as the direction of force and motion are not identical.

Using Newton's law we can write :

The frictional force is given by :

Its direction will be opposite of the direction of the motion. Thus acceleration produced will be negative.

Thus the net acceleration is = 3.5 - 0.98 = 2.52 m/s^{2}.

The total distance travelled is given by :

or

Hence the work done by applied force is given by :

The work done by frictional force will be negative as the force opposes the motion.

The net work done will be the sum of work done by applied force and work done by frictional force.

It is given that initial velocity is zero. The final velocity can be calculated by the equation of motion :

or

or

Thus change is kinetic energy is :

or

or

Total energy = kinetic energy (KE) + potential energy(PE)

**KE > 0 since m and v ^{2} is positive. If KE <0 particles cannot be find. If PE>TE, then KE<0 (now in all graph check for this condition)**

**In case 1** kinetic energy is negative for x<a. So at x<a particle cannot be found.

**In case 2 **for x<a and for x> b kinetic energy is negative. So the particle cannot be found in these regions.

**In the third case**, the minimum potential energy is when . At this position, the potential energy is negative (- V_{1}).

The kinetic energy in this case is given by :

And the minimum energy of particle is - V_{1}.

**In the fourth case**, the particle will not exist in the states which will have potential energy greater than the total energy.

Thus particle will not exist in and .

The minimum energy of particle will be - V_{1} as it is the minimum potential energy.

reaches

The total energy of the particle is given by :

or

At the extreme position, the velocity of the object is zero thus its kinetic energy at that point is zero.

or

or

or

Hence the extreme position are .

The total energy is given by :

or

The burning of casing results in a reduction in the mass of the rocket. This leads to a lowering in the total energy.

Thus heat required for burning is obtained from the expenses of the **rocket**.

This is because the gravitational force is a conservative force. And we know that the work done by a conservative force in a closed path is always zero. That's why the work done by the gravitational force is zero in a complete orbit revolution.

The total energy of artificial satellite remains constant. Thus when it approaches towards the earth the distance between them decreases. This results in a decrease in the potential energy of the satellite. By energy conservation, the kinetic energy of satellite increases and so does the velocity.

In the first case,

Work done is :

or

or

or

In the second case :

or

or

Thus work done in the second case is greater than the first case.

It is given that work done by the conservative force is positive, thus the force acts in the direction of the motion. This results in a decrease in distance between the bodies. Thus it's potential energy **decreases**.

**Q6 (b)** Work done by a body against friction always results in a loss of its kinetic/potential energy.

Work done by the body against friction results in a decrease in the velocity of the body. Thus the kinetic energy of the body decreases.

The internal force cannot produce a change in the total momentum as no external force is acting. Thus the change in total momentum is proportional to the external forces acting on the body.

The conservation of **total linear momentum** doesn't depend upon the fact whether it is an elastic collision or an inelastic collision.

** Q7(a) **State if each of the following statements is true or false. Give reasons for your answer.

In an elastic collision of two bodies, the momentum and energy of each body is conserved.

**False:**- The linear momentum and energy will be conserved if both are considered in a system. But for individual bodies, this conservation of momentum and energy doesn't hold. This is because the impact during the collision may transfer energy/momentum of one ball to the other ball.

**False:**- Internal forces will not change the energy of the system but external forces can change the total energy by changing their magnitude or direction.

**False:-** This is true only for conservative forces e.g. gravitational force. For e.g in case of frictional force (non-conservative force), the work done in a closed-loop cannot be zero as energy is wasted throughout.

**True but not always:- ** In the case of inelastic collisions, few amounts of energy is converted into other forms of energy such as sound or in deformation. Thus final kinetic energy is always less as compared to initial kinetic energy. But in case of the explosion of a bomb final kinetic energy is greater than the initial kinetic energy

No, because at the time of the collision, the kinetic energy is converted to the potential energy. Thus total kinetic energy is not constant at the collision.

** Q8 (b) ** Is the total linear momentum conserved during the short time of an elastic collision of two balls?

Yes, in case of elastic collision the total linear momentum of the system remains conserved as no external force is acting on the system of balls.

The total kinetic energy of the system cannot be conserved in case of inelastic collision as there is loss of energy in the form of deformation. But the total linear momentum of the system remains constant even in the case of inelastic collision as no external force is acting.

Since the potential energy of the system depends upon the separation between the bodies thus the forces acting on the body are conservative in nature. We know that conservative forces produce elastic collisions.

It is given that acceleration is constant thus force will also be constant (by Newton's law of motion F = ma).

Also,

or

Thus

Now, the work done by the force is given by :

Hence power is directly proportional to the time.

We know that the power is given by :

or

or

It is given that power is constant, thus :

or

By integrating both sides, we get

Also, we can write :

or

By integrating we get the relation :

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?

Force is given to be :

And the displacement is :

Thus the work done is given by :

or

or

)

The kinetic energy of the electron is given by :

or

Thus velocity is obtained as :

or

Similarly, we can find the velocity of the proton :

Thus velocity is obtained as :

or

Thus the ratio of their velocities is :

The volume of the drop is :

Thus the mass of raindrop is :

or

Thus the work done is given by :

or

or

or

Now the total energy at the peak point is :

or

or

And the energy at the ground is :

or

or

Thus work done by the resistive force is :

The momentum is conserved in the collision as no external force is acting on the system. In the given case the rebound velocity is the same as the initial velocity thus the kinetic energy of the molecule initially and finally are same. Hence this is an elastic collision.

Mass of the water is :

or

Thus the output power is given by :

or

or

or

Also, we are given that efficiency is 30 per cent.

Thus the input power is :

or

The initial kinetic energy of the system is given by :

or

**Case (i):-** The final kinetic energy is :

Thus the kinetic energy is not conserved in this case.

**Case (ii):- **The final kinetic energy is :

**Thus kinetic energy is conserved in this case.**

**Case (iii):- **The final kinetic energy is:-

or

Thus the kinetic energy is not conserved in this case.

on a table as shown in Fig. 6.15. How high does the bob A rise after the collision? Neglect the size of

the bobs and assume the collision to be elastic.

This is an elastic collision thus the transfer of momentum will take place. It is given that bob B is at rest and bob A has some velocity. So in momentum transfer, bob B will gain the velocity in the left direction whereas bob A will come to rest (complete momentum transfer takes place).

Hence bob A will not rise.

Consider the extreme position (horizintal) :-

The kinetic energy at this position is zero as velocity is zero.

Thus total energy is given by :

Now consider the mean position (lowermost point) :

Here the potential energy of bob is zero.

Whereas kinetic energy is :

Further, it is given that 5 per cent of energy is dissipated due to air resistance while coming down.

Thus energy equation becomes (conservation of energy):-

or

Since the sand is falling in the trolley thus the force generated on the system (trolley and sandbag) is an internal force. There is no external force thus momentum of the system doesn't change. Hence speed remains the same i.e., 27 Km/hr.

The relation between work done and the kinetic energy is given by :

Using the relation we can write :

Initial velocity = 0 (at x = 0 )

And the final velocity = (at x = 2).

Thus work done is :

or

or

The volume of wind = here is the swept circle and is the velocity.

Thus the mass of the wind is : - _{} , is the density of the air.

Hence mass of wind flowing through windmill in time t is _{.}

**Q21 (b) ** The blades of a windmill sweep out a circle of area A. What is the kinetic energy of the air?

The kinetic energy is given by :

or

or

Thus the kinetic energy of wind is _{} J.

It is given that 25 per cent of wind energy is converted into electrical energy.

Thus electric energy produced is :

or

Now the electric power is given by :

or

or

or

The work done against the gravitational force is given by :

= Number of times the weight is lifted work done in 1 time.

or

or

Efficiency is given to be 20 per cent.

Thus energy supplied by the person :

Thus the amount of fat lost is :

or

It is given that the efficiency of energy conversion is 20 per cent.

According to question, we can write (equating power used by family) :

(Here A is the area required.)

or

or

Thus required area is 200 m^{2}.

**Q23 (b) ** A family uses 8 kW of power. Compare this area to that of the roof of a typical house.

A typical has dimensions of .

The area of the roof of the house is .

This is nearly equal to the area required for the production of the given amount of electricity.

We are given :

Mass of the bullet m: 0.012 Kg

Mass of the block M: 0.4 Kg

The initial velocity of the bullet u: 70 m/s

The initial velocity of the block : 0

The final velocity of the system (bullet + block): v

For finding the final speed of system we will apply the law of conservation of momentum :

or

Now for the system, we will apply the law of conservation of energy :

The potential energy at the highest point = Kinetic energy at the lowest point

or

or

or

Hence heat produced is :

or

or

The FBD of the track is shown in the figure below :

Using the law of conservation of energy we have :

or

Hence both stones will reach the bottom with the same speed.

**For stone 1 **we can write :

or

**For stone 2 **we have :

Also, using the equation of motion,

or

It is given that

or

Thus

Hence, the stone travelling on the steep plane will reach before.

For finding speed and time we can use conservation of energy.

or

or

or

And the time is given by :

and

Displacement (x) of the block is given as : = 0.1 m.

Using equilibrium conditions we can write :

and ( _{} is the frictional force).

We can write work done in terms of potential energy as :

or

or .

Thus the coefficient of friction is 0.125.

In this case, the heat produced is the loss in the potential energy.

Thus,

heat produced = mg h

or

or

The heat produced (when the lift is stationary) will remain the same as the relative velocity of the bolt with respect lift still remains zero.

The initial momentum of the system (boy + trolley) is given as :

Now assume *v*' is the final velocity of the trolley with respect to the ground.

Then the final momentum will be :

Conserving momentum :

or

The time taken by the boy is :

Hence the distance moved by the trolley is :

The potential energy of the system depends inversely on the separation between the balls. Thus the potential energy will decrease as the balls will come closer and will become zero as they touch each other.

Thus elastic collision is best described only by the graph (v).

By Einstein’s mass-energy relation we can write :

Here and C are constant thus two-body decay is unable to explain (or account for) the continuous energy distribution in the *β*-decay of a neutron.

- Many questions were done in the solutions of NCERT class 11 physics chapter 6 work energy and power based on the concept of the work-energy theorem which is important throughout the physics course. The concept will be used in class 12 also.
- CBSE NCERT solutions for class 11 physics chapter 6 work energy and power will help to score well in-class exams and also competitive exams like NEET.
- Learning NCERT solutions will help to solve problems from other reference books also.

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