Work, Energy And Power

From work energy theorem, we know that change in kinetic energy of the system is equal to work done by net force acting on the system. According to the question, the magnetic forces due to motion of an electron and a proton act in perpendicular direction to the direction of motion which results in the centripetal force for the particle and hence particle performs uniform circular motion (i.e.; speed of the particle remains constant)

∴ no work is done by the forces and so no change in K.E. of the particle.

Hence, magnetic forces do no work on each particle.

Option (b) is correct.

We know that, force between two protons is equal to the force

between proton and a positron (∵ their charges are same)

now,

mass of positron = times mass of the proton

greater the mass lesser the distance a particle travels

∴ positron will travel the largest distance

And ,work done=

Force is same in both the cases only distance varies (positron’s

Distance is larger than proton) and

∴work done will be more in case of positron.

Option (c) is correct.

According to the question more force is required by the man

during the process of standing than when he stands up i.e.;

during standing up, normal force = friction + mg ⇒ N>mg

while standing N=mg

option (d) is correct.

We know that, Work done=F× d

But here road is not moving ∴ work done by cycle on

road is equal to zero.

Option (c) is correct.

During free fall,

K.E. increases, momentum increases and P.E. decreases

Total mechanical energy=K.E.+P.E.

As P.E. decreases and K.E. increases ∴ M.E. remains constant.

Option (c) is correct.

In inelastic collision no external force is present.

∴ total linear momentum of the system remains conserved.

Option (c) is correct.

kinetic energy appears in various forms i.e.; energy

May be lost in various forms of heat, sound etc.

∴ either (initial K.E. or final K.E. will be greater)

M.E.=K.E.+P.E. (∵ K.E. Is not constant ∴ M.E. will also be

not constant)

Option (a), (b), (d) is incorrect.

Considering the ball in incline AB has mass ,acceleration

and the ball in incline AC has mass ,acceleration .

from the above fig.

As,

We know that ,

From above equation we see that t is inversely proportional to a.

∴

Now,

From law of conservation of energy, we get to see in the above

Fig. that

K.E. at B=K.E. at C =P.E. at A

∴ ….(1)

…. (2)

Option (c) is correct.

Given, force constant k=0.5N/m.

V(x) =

k.E. =

from law of conservation of energy,

total energy=K.E. + P.E.

At the moment of turns back then velocity=0⇒ K.E.=0

∴ total energy =P.E.

option (b) is correct.

for elastic collision initial K.E. = final K.E.

let m be the mass of each ball bearings

K.E. before collision =

Now, K.E after collision in different cases

Case(a):

Case (b):

Case (c):

Case (d):

From the above cases we see that in case(b) K.E. is conserved.

Option (b) is correct.

Given,

Initial velocity, u (at x=0) =0

Final velocity ,v(at x=2m)=

Work done=change in kinetic velocity

=

=

=

Option (b) is correct.

According to the question, body is moving unidirectionally

under the influence of a source of constant power supplying

energy. We know that,

power=force× velocity

P=F× v

P=ma× v…… (1)

We also know that, v=u+at…. (2)

V=at (∵ u=0 body starts initially from rest)

From eq (1) and (2)

P=ma× at

P =

… (3)

We also know that,

(∵ u=o)….(4)

From eq (3) and (4)

⇒

From the above equation we see that s is directly proportional to

Displacement-time graph is correctly shown by Option (b).

when earth is near the sun then due to gravitational pull

K.E. decreases and when earth is away from sun gravitational

pull is less as a result K.E. increases.

option(d) best represents it.

When a pendulum oscillates in air its total mechanical energy

decreases continuously in overcoming resistance due to air.

We know that, total mechanical energy of pendulum decreases

exponentially with time

∴ option (c) is correct representation

of it.

Given, m=5kg, r=1m, frequency of revolution, f =300re/min

Kinetic energy=

∴ angular velocity =

=

= (300× 2× 3.14) rad/60s

= rad/s

=

We know that v=wR

=

=

Kinetic energy=

=

we know that at height h, P.E. is maximum and K.E.=0(∵ v=0)

Now, as the raindrop falls its P.E. starts decreasing and K.E.

starts increasing. Hence, total mechanical energy remains

conserved if air resistance is neglected.

If there is some air resistance, then there exists a force called

upthrust. So, as velocity increases upthrust also increases.

∴ during rainfall first its velocity increases then become

constant after sometime. (this velocity is termed as terminal

velocity.) Hence, K.E. also becomes constant and P.E.

decreases continuously as drop is falling continuously.

option (b) represents the correct diagram.

Given, h=1.5m, v=1m/s , a= ,m=10kg

If air resistance is negligible then total mechanical energy

remains constant (P.E. at ground is zero)

initial energy=

=

=

=155J

Now,

from law of conservation of mechanical energy,

⇒

∴ final K.E.=155J.

option (d) is correct.

When an iron sphere is falling freely in the lake, it will

accelerate first then its velocity increases and resistance

due to water causes a viscous force which oppose its

motion. Hence, during fall of the sphere, first its velocity

increases and then become constant after sometime.

option (b) is best representable graph of it.

Given, m=150g=0.15kg, u=126km/hr=35m/s, ∆t=0.001s

v=-35m/s

it is an elastic collision.

we know that

(∵ u=v,v=-v)

option (c) is correct.

according to question,

When a man of mass m ,climbs up the staircase of height L

work done by gravitational force on man =-mgL

work done by muscular forces=- (work done against gravitational

force) =mgL

∴ work done by all the forces =mgL-mgL=0

as the point of application of contact forces does not move

so, work done is zero.

option (b, d) is correct.

From law of conservation of energy,

⇒

⇒

⇒ …..(1)

it is given that while it is moving downward it emerges out with

half the kinetic energy

⇒

⇒ (substituting the value of v’ from eq (1))…..(2)

from eq (1) and (2)

⇒

∴ velocity of the bullet is more than half of its earlier velocity.

Hence, the bullet will move in a different parabolic path and internal

energy of the particle will increase.

option (b,d,f) is correct

Collision is elastic ∴ velocities gets interchanged and K.E. and

linear momentum remains conserved.

when comes in contact with spring, it is retarded by the

spring force and is accelerated by the spring force.

now let’s see the cases,

(a) the spring will compress until two blocks acquire common

velocities .some of the K.E. of is stored as P.E. and

some into K.E. of ∴ (a) is incorrect.

(b)momentum is conserved. option (b) is incorrect.

(c) In head on elastic collision two bodies exchange their

velocities and so finally will come to rest.

option (c) is correct.

(d) ∵ there is loss in K.E. when the blocks collide on rough

surface therefore, collision is inelastic. Option (d) is correct.

There is no dissipation of energy as the block rests on the

incline, the force of friction acting between block and incline

opposes the block to slide and as a result the block is not

in motion with respect to incline therefore, work done by

force of friction between block and incline plane is zero.

Electrical power is required when an elevator is descending

to prevent it from free fall under gravity.

The cable of the elevator has the capacity of some limiting

value of tension developed in it. therefore, there is a limit

on the number of passengers.

(a) When force is applied to raise the body then the angle

between force and displacement is 0°.

∴ Work done =Fscosθ =Fscos0=Fs

sign of work done is positive.

(b)As gravitational force acts in downward direction then the

angle between force and displacement is 180°.

∴ work done =Fscosθ =Fscos180°=-Fs.

sign of work done is negative.

Given m=400kg, d=2m

car is displaced horizontally and force (weight of the car) is

acting downward, angle between force and displacement is 90 �

∴ Work done=FscosƟ=Fscos90 �=0

Thus, work done is zero.

As we know that gravitation is conservative force;

Therefore, mechanical energy of the body falling towards

earth in air will be conserved.

No, the work done in moving the body is not necessarily zero until the body reaches its initial position. If the body does not reach the

During the collision the there might occur deformation in the billiard balls due to which some of the kinetic energy may get lost in the potential energy. While linear momentum is always conserved so only linear momentum will be conserved at the time of collision.

Given: mass = 100kg

Initial Height = 0 m

Final height = 10 m

Time taken t = 20 s

Work done in lifting the mass W

= change potential energy

=

∴ Power = Work done per unit time =

Work done by heart in one beat = 0.5 J

∵ it beats 72 times in a minute

∴ Work done in minute =

∴ Power used by heart = work done per unit time =

Imagine a pendulum doing its oscillatory motion. Now the tension due to the rod is always perpendicular to the displacement. ∴ the work done by the tension force will always be the zero and hence it will never bring about any change in kinetic energy.

For particle a,

Change in kinetic energy = work done by the force

For particle b,

Change in kinetic energy = work done by the force

Now,

∵ and the magnitude of the force is same

∴ displacement will also be the same

(a) Vertically downward as the tangential velocity at the point B will be vertically downwards

(b) Parabolic path with vertex at C as the tangential velocity will be in the horizontal direction but due to downward gravitational force it will follow parabolic path

(c)It will also be a parabolic path with vertex at x, due to the tangential velocity and the gravitational force

i.) Kinetic energy vs x

We know that,

At C, P.E =0,

∴

Till point D there is no change in P.E, ∴ the K.E remains constant

At F,

∴

From B to C the potential decreases so the kinetic energy increases ∴ we get inverted curve of potential.

From C to D potential remains constant so the kinetic energy will also remain same and thus we get a straight line parallel to x- axis

From D to F the potential energy increases to E_o linearly so the kinetic energy must fall linearly to E_o.

∴ we get the following plot,

ii.) Velocity vs X

Now,

-----(2)

From B to C, from above equation, velocity increases parabolically symmetric from x-axis.

From C to D there is no change in P.E energy so the velocity remains constant.

From D to F, the P.E increases linearly and thus the velocity decreases parabolically from equation (2)

Now since we get two values of velocities from equation (2), we will get a mirror image of the graph in the negative side of the fourth quadrant too.

∴ we obtain following the graph,

a) In this problem if we prove that both the velocities after collision is positive will be enough to prove part a)

Conserving momentum,

---(1)

Where are the velocity of both the ball

Putting equation 1 in above equation we get,

from this we get,

∵

For first inequality, ------(2)

From second inequality, -------(3)

Combining (2) and (3),

We get,

∵ after collision both the velocities are positive

∴ both the balls move in same forward direction.

b)

Let p_o be initial momentum and p₁and p₂be the momentum of ball 1 and 2 respectively after collision. (here bold text means vector)

In inelastic collision,

Conserving momentum,

By vector addition of momentums, we get,

-----(1)

But ∵ there is a loss in kinetic energy

∴

----- (2)

From equation (1) we can write equation (2) as,

-------- (3)

∴ the angle between the two velocities of scattered balls is less than 90°.

(Note: this problem is based the concept that potential energy can be negative also and E = V+K)

Region A: No, it is not possible

∵ and

∴ which is not possible

Region B: Yes, it is possible.

∵ and

∴ which is true

Region C: Yes, it is possible.

∵ and

∴ which is possible

Region D: Yes, it is possible.

∵ which is possible

Let be the mass of the bob and be the initial height of the bob.

Potential energy of a particle raised to the height h is =

And kinetic energy of the particle with velocity v is given as

=

Just before collision, bob A reaches ground and its potential energy gets converted to kinetic energy,

∴ conserving energy we get,

-----(1)

(a) Using the result that when two bodies of same mass collide, their velocities are mutually interchanged,

Bob A must come to rest and bob B must gain the velocity in equation (1).

a) ∴ from above we conclude the bob A will not rise to any height since its velocity will become zero after collision.

b) ∵ the velocities are interchanged,

∴ velocity of bob B =

Mass m of drop = 1g = 0.001 kg

Change in height Δ h = 1 km = 1000 m

Final velocity = 50

∵ the drop starts from rest, initial velocity = 0

∴ initial K.E = 0 J

And final K.E =

a) Loss in P.E. =

b) Gain in K.E.

c) No, the gain in K.E. equal to loss of P.E. is not the same because of resistance like drag and viscous force against the drop.

a) At time period t=0

Bob A has maximum potential energy and bob B is at rest with zero energy.

At t=T/4

Bob A gains maximum velocity and collides with the bob B. bob A comes to rest and bob B gains the velocity of bob A because the collision is elastic and both have the same mass ∴ their velocity gets interchanged.

Now bob B has maximum kinetic energy.

At t=T/2

Bob B reaches the maximum height and its kinetic energy is converted to maximum potential energy. While bob A remains at rest.

At =3T/4

Bob B reaches to bob A with its maximum potential energy getting converted to max kinetic energy and collides with bob A. During the collision the interchanging of velocity takes place and bob A gains the velocity of bob B and bob B comes to rest.

At t=T

Bob A reaches the maximum height and its kinetic energy is converted to maximum potential energy. While bob B remains at rest.

And after t=T the whole process between 0<t<T repeats itself in the time period of T<t<2T

b) Here E is total energy of bob A and B ∴ we get straight lines, since they E is constant.

Now from the above discussion in part a) we get the following graphs,

For bob A,

For bob B,

Average mass of the raindrop =

Terminal velocity v

Kinetic energy before hitting the ground

∴ energy transferred by 1 drop

Volume of water received =

∴ total mass of the water received

∴ no. of droplets =

∴ total energy transferred by all the droplets

Mass of car m = 50,000 kg

Velocity v

∴ kinetic energy

∵ 90% of the energy is lost due to friction

∴ only 10% of the energy is stored by the spring

∴ where k is spring constant and x=compression

1 step is of 1 m

∴ in 6km there are 6000 steps

Person raises the body by 0.25 m in each step

Work done in each step =

∴ total work in 6000 steps

W=

W = 10% of food energy

Food energy = 10× W=

In 1litre of the petrol car moves distance d=15km=15000 m

With 0.5 efficiency heat energy to drive the car in one litre

Since the car moves with the uniform velocity,

Work done by the force of friction = heat energy produced by 1litre of petrol

∴ the force of friction = 1000 N

Mass of the block m = 1kg

Angle of inclination =

Coefficient of friction = 0.1

Displacement = 10 m

a) Work done against gravity

=

b) Work done against friction

=

c) Vertical raise in height h

∴ change in P.E.

d) Applying work energy theorem,

-(1)

Putting these in equation (1),

We get,

∴ there is a decrease in the kinetic energy by 41.3 J

e) Applied force F =

Displacement d =

Work done W by the force F is given as

=

=

(a) ball 1 and 3 the total mechanical energy is conserved.

For ball 1, friction is so large that starts rolling without slipping and gains rotational kinetic energy.

For ball 3, friction is negligible so there is no external force responsible for energy loss ∴ the energy is conserved.

(b) ball 1 acquires some rotational kinetic energy ∴ it will not be

able to reach C and thus will not reach point D

ball 2 loses its energy due to friction, thus it cannot reach point D

ball 3 has exactly the amount of energy to reach C and D because it does not roll and does not lose energy due to friction.

∴ only ball 3 will reach point D

(c) ball 1 and 2 starts moving in the back direction before

reaching point C.

for ball 1, it loses its energy while slipping back to point B, thus it will not reach back to point A

for ball 2, it will have clockwise rotation as it moves backs to point B due to which there will be kinetic friction coming to play due to which it will lose energy and won’t be able to reach point A.

At time t,

Kinetic energy

Now at time t+Δt,

Rocket loses mass Δm and gains velocity Δv and releases gas with relative velocity of u

Velocity of rocket w.r.t ground

Velocity of gas w.r.t ground

∴ total kinetic energy of the system w.r.t ground

[neglecting terms with as both of them are very small]

Kinetic gained in Δt interval,

----(1)

∵ there is no external force ∴ internal forces must cancel out,

Putting the above in equation (1), we get,

By work energy theorem,

Work done = change in kinetic energy

W

∴ the device that ejects gas does work = ∆m u² in this time interval

From Hooke’s law, ,

Where F = Force,

A= Area of cross section

Y= young’s modulus

L=initial length of the cube

Δ L=change in length

Kinetic energy of both boxes K.E.

Now, the kinetic energy gets converted to potential energy due to compression and,

P.E.

Now, P.E = K.E

According to Archimedes principle when a body is partially or fully dipped into a fluid at rest, the fluid exerts an upward force of buoyancy equal to the weight of the displaced fluid.

The balloon similarly experiences a buoyant force which provides it an acceleration .

If

are the mass , density and volume of the balloon andis the density of

.

Then,

Now writing acceleration a, in its differential form,

Integrating both the sides, we obtain,

Now for the height by which it raises till time t, we use equation of motion

Putting the value of a and initial velocity to be zero,

now the kinetic energy of the balloon at time t,

K.E

Rearranging the terms, we get,

But are terms which represent potential energy of air and helium balloon respectively.

∴

So, we can say the balloon raises by pushing the air downward, thus its potential energy increases as the air’s decreases.