Work Power & Energy — Page 4 | JEE Physics

Q31

Given below are two statements: Statement I : A truck and a car moving with same kinetic energy are brought to rest by applying breaks which provide equal retarding forces. Both come to rest in equal distance. Statement II : A car moving towards east takes a turn and moves towards north, the speed remains unchanged. The acceleration of the car is zero. In the light of given statements, choose the most appropriate answer from the options given below

A Statement I is incorrect but Statement II is correct.

B Statement

\mathrm{I}

is correct but Statement II is incorrect.

C Both Statement I and Statement II are correct.

D Both Statement I and Statement II are incorrect.

Correct Answer

Option B

Solution

Statement I is correct: The kinetic energy of an object is given by $\dfrac{1}{2}mv^2$ , where m is the mass of the object and v is its velocity.

If a truck and a car are moving with the same kinetic energy and are brought to rest by applying brakes that provide equal retarding forces, both will come to rest in equal distances.

This is because the distance required to stop an object depends on its initial kinetic energy and the force applied to bring it to rest.

Since both the truck and car have the same initial kinetic energy and are subjected to the same retarding force, they will come to rest in the same distance.

Statement II is incorrect.

When the car moves from east to north, even though its speed remains unchanged, its direction changes.

Since velocity is a vector quantity that has both magnitude (speed) and direction, a change in direction implies a change in velocity.

Acceleration is the rate of change of velocity, so when the velocity changes, there is acceleration.

In this case, the car's acceleration is not zero as it turns from east to north.

Q32

A ball suspended by a thread swings in a vertical plane so that its magnitude of acceleration in the extreme position and lowest position are equal. The angle

(\theta)

of thread deflection in the extreme position will be :

A

\tan ^{-1}\left(\dfrac{1}{2}\right)

B

2 \tan ^{-1}\left(\dfrac{1}{2}\right)

C

2 \tan ^{-1}\left(\dfrac{1}{\sqrt{5}}\right)

D

\tan ^{-1}(\sqrt{2})

Correct Answer

Option B

Solution

Loss in kinetic energy $=$ Gain in potential energy

\begin{aligned} & \Rightarrow \frac{1}{2} \mathrm{mv}^2=\mathrm{mg} \ell(1-\cos \theta) \\ & \Rightarrow \frac{\mathrm{v}^2}{\ell}=2 \mathrm{~g}(1-\cos \theta) \end{aligned}

Acceleration at lowest point

=\frac{\mathrm{v}^2}{\ell}

Acceleration at extreme point

=g \sin \theta

\begin{aligned} & \text{ Hence, } \frac{\mathrm{v}^2}{\ell}=\mathrm{g} \sin \theta \\ & \therefore \sin \theta=2(1-\cos \theta) \\ & \Rightarrow \tan \frac{\theta}{2}=\frac{1}{2} \Rightarrow \theta=2 \tan ^{-1}\left(\frac{1}{2}\right) \end{aligned}

Q33

A block of mass

100 \mathrm{~kg}

slides over a distance of

10 \mathrm{~m}

on a horizontal surface. If the co-efficient of friction between the surfaces is 0.4, then the work done against friction

(\operatorname{in} J

) is :

A 3900

B 4500

C 4200

D 4000

Correct Answer

Option D

Solution

\begin{aligned} & \text{ Given } \mathrm{m}=100 \mathrm{~kg} \\ & \mathrm{~s}=10 \mathrm{~m} \\ & \mu=0.4 \\ & \text{ As } \mathrm{f}=\mu \mathrm{mg}=0.4 \times 100 \times 10=400 \mathrm{~N} \\ & \text{ Now } \mathrm{W}=\mathrm{f} . \mathrm{s}=400 \times 10=4000 \mathrm{~J} \end{aligned}

Q34

If a rubber ball falls from a height

h

and rebounds upto the height of

h / 2

. The percentage loss of total energy of the initial system as well as velocity ball before it strikes the ground, respectively, are :

A

50 \%, \sqrt{2 \mathrm{gh}}

B

50 \%, \sqrt{\mathrm{gh}}

C

50 \%, \sqrt{\dfrac{\text{ gh }}{2}}

D

40 \%, \sqrt{2 \mathrm{gh}}

Correct Answer

Option A

Solution

To solve this problem, we need to analyze both the energy loss and the initial velocity of the rubber ball before it strikes the ground.

First, let's consider the energy loss.

The energy involved here is gravitational potential energy.

The initial potential energy of the ball when it is about to fall is given by

U_i = mgh

, where

U_i

is the initial potential energy,

m

is the mass of the ball,

g

is the acceleration due to gravity, and

h

is the initial height from which the ball falls. After the ball rebounds, it reaches a height of

h/2

. The potential energy at this new height is

U_f = mg \cdot \frac{h}{2}

.

The energy loss can be calculated as the difference between the initial and final potential energies, and to find the percentage energy loss, we divide this difference by the initial energy and multiply by 100:

\text{Energy loss percentage} = \frac{(U_i - U_f)}{U_i} \times 100

Substituting the values of

U_i

and

U_f

gives:

\text{Energy loss percentage} = \frac{(mgh - mg\frac{h}{2})}{mgh} \times 100

By simplifying, we find:

\text{Energy loss percentage} = \frac{mgh - \frac{1}{2} mgh}{mgh} \times 100 = \frac{1}{2} \times 100 = 50\%

This tells us that the energy loss percentage is indeed

50\%

.

Next, we'll find the velocity of the ball just before it strikes the ground.

The velocity can be determined using the formula for the velocity of an object in free fall:

v = \sqrt{2gh}

Here,

v

is the velocity of the ball just before impact,

g

is the acceleration due to gravity, and

h

is the height from which the ball falls.

This formula shows that the initial velocity of the ball before it strikes the ground is

\sqrt{2gh}

, not taking into account air resistance and assuming it starts from rest. Therefore, the correct answer is Option A:

50\%

,

\sqrt{2gh}

.

Q35

When kinetic energy of a body becomes 36 times of its original value, the percentage increase in the momentum of the body will be :

A 60%

B 500%

C 6%

D 600%

Correct Answer

Option B

Solution

The relationship between kinetic energy (K.E) and momentum (p) of a body can be expressed through their respective definitions.

Kinetic energy is given by

K.E = \frac{1}{2} mv^2

where $m$ is the mass of the body and $v$ is its velocity. The momentum (p) of a body is given by

p = mv

. To express kinetic energy in terms of momentum, we can manipulate the expression for momentum as follows:

p = mv \implies v = \frac{p}{m}

Substituting $v$ in the kinetic energy formula, we get

K.E = \frac{1}{2} m\left(\frac{p}{m}\right)^2 = \frac{1}{2} \frac{p^2}{m}

Therefore, we see that kinetic energy is directly proportional to the square of the momentum $(K.E \propto p^2)$ .

Now, given that the kinetic energy of a body becomes 36 times its original value, we can set up the proportionality as

\frac{K.E_{\text{final}}}{K.E_{\text{original}}} = 36

Since $K.E_{\text{final}} = 36 \times K.E_{\text{original}}$ and knowing $K.E \propto p^2$ , we can express this relationship through the squares of the initial and final momentum:

\frac{p_{\text{final}}^2}{p_{\text{original}}^2} = 36

Taking the square root of both sides to find the ratio of final to initial momentum, we have

\frac{p_{\text{final}}}{p_{\text{original}}} = \sqrt{36} = 6

This indicates that the final momentum is 6 times the original momentum.

To find the percentage increase in the momentum, we calculate the increase from the original to the final, subtracting the original momentum (which is considered 1 times itself):

\text{Percentage increase} = \left(\frac{p_{\text{final}} - p_{\text{original}}}{p_{\text{original}}}\right) \times 100\% = \left(\frac{6p - p}{p}\right) \times 100\% = \left(6 - 1\right) \times 100\% = 5 \times 100\% = 500\%

Therefore, the correct answer is Option B: 500%.

Q36

A bullet of mass

50 \mathrm{~g}

is fired with a speed

100 \mathrm{~m} / \mathrm{s}

on a plywood and emerges with

40 \mathrm{~m} / \mathrm{s}

. The percentage loss of kinetic energy is :

A

44 \%

B

16 \%

C

84 \%

D

32 \%

Correct Answer

Option C

Solution

To find the percentage loss of kinetic energy of the bullet, we first calculate the initial kinetic energy before the bullet hits the plywood and the final kinetic energy after it emerges.

The formula for kinetic energy (KE) is given by:

KE = \frac{1}{2} mv^2

where $m$ is the mass of the object and $v$ is its velocity.

Let's calculate the initial and final kinetic energies.

Initial Kinetic Energy:

KE_{\text{initial}} = \frac{1}{2} \times 50 \times (100)^2 = \frac{1}{2} \times 50 \times 10000 = 25 \times 10000 = 250000 \, \text{g.m}^2/\text{s}^2

Note: To keep units consistent, we used grams and meters per second.

We can also convert the mass to kilograms (by dividing by 1000) which would result in the energy being calculated in Joules, but for the purpose of finding the percentage change, the form of units does not matter as long as they are consistent, since it will be a ratio.

Final Kinetic Energy:

KE_{\text{final}} = \frac{1}{2} \times 50 \times (40)^2 = \frac{1}{2} \times 50 \times 1600 = 25 \times 1600 = 40000 \, \text{g.m}^2/\text{s}^2

The loss of kinetic energy is then:

\Delta KE = KE_{\text{initial}} - KE_{\text{final}} = 250000 - 40000 = 210000 \, \text{g.m}^2/\text{s}^2

Finally, the percentage loss of kinetic energy can be calculated using the formula:

\text{Percentage loss of KE} = \left( \frac{\Delta KE}{KE_{\text{initial}}} \right) \times 100\%

\text{Percentage loss of KE} = \left( \frac{210000}{250000} \right) \times 100\% = 0.84 \times 100\% = 84\%

Thus, the percentage loss of kinetic energy is 84%, which corresponds to Option C.

Q37

A body of mass 4 kg is placed on a plane at a point

P

having coordinate

(3,4) \mathrm{m}

. Under the action of force

\overrightarrow{\mathrm{F}}=(2 \hat{i}+3 \hat{j}) \mathrm{N}

, it moves to a new point Q having coordinates

(6,10) \mathrm{m}

in 4 sec . The average power and instanteous power at the end of 4 sec are in the ratio of :

A 4 : 3

B 13 : 6

C 1 : 2

D 6 : 13

Correct Answer

Option D

Solution

\begin{aligned} & \langle\mathrm{p}\rangle=\frac{(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}) \cdot(3 \hat{\mathrm{i}}+6 \hat{\mathrm{j}})}{4}=6 \\ & \overrightarrow{\mathrm{a}}=\left(\frac{\overrightarrow{\mathrm{F}}}{\mathrm{~m}}=\frac{1}{2} \hat{\mathrm{i}}+\frac{3}{4} \hat{\mathrm{j}}\right) \\ & \overrightarrow{\mathrm{v}} \text{ at } \mathrm{t}=4 \mathrm{sec}=\left(\frac{1}{2} \hat{\mathrm{i}}+\frac{3}{4} \hat{\mathrm{j}}\right) \times 4=(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}) \\ & P_{\text{ins }}=(2 \hat{\mathrm{i}}+3)(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}})=13 \\ & \frac{\langle\mathrm{P}\rangle}{P_{\text{ins }}}=\frac{6}{13} \end{aligned}

Note: Given data is not matching.

\begin{aligned} & S=u t+\frac{1}{2} a t^2 \\ & S=0+\frac{1}{2} \frac{(2 \hat{i}+3 \hat{j})}{4}(4)^2=4 \hat{i}+6 \hat{\mathrm{j}} \end{aligned}

If $\overrightarrow{\mathrm{r}}_{\mathrm{i}}=3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}$ then $\overrightarrow{\mathrm{r}}_{\mathrm{f}}=7 \hat{\mathrm{i}}+10 \hat{\mathrm{j}}$ But Final position given in the question is $(6,10)$ .

Q38

A ball having kinetic energy KE, is projected at an angle of

60^{\circ}

from the horizontal. What will be the kinetic energy of ball at the highest point of its flight?

A

\dfrac{(\mathrm{KE})}{2}

B

\dfrac{(\mathrm{KE})}{8}

C

\dfrac{(\mathrm{KE})}{4}

D

\dfrac{(\mathrm{KE})}{16}

Correct Answer

Option C

Solution

Let's break down the problem step by step: The ball is projected with kinetic energy

\mathrm{KE} = \frac{1}{2} m v^2,

where

v

is the initial speed. The ball is launched at an angle of

60^\circ

. So, its initial velocity components are: Horizontal:

v_x = v \cos(60^\circ) = \frac{v}{2}

Vertical:

v_y = v \sin(60^\circ)

At the highest point of its flight, the vertical component of the velocity becomes zero (i.e.,

v_y = 0

), but the horizontal component remains unchanged.

Therefore, the kinetic energy at the highest point is only due to the horizontal velocity:

\mathrm{KE}_{\text{top}} = \frac{1}{2} m v_x^2

Substitute

v_x = \frac{v}{2}

:

\mathrm{KE}_{\text{top}} = \frac{1}{2} m \left(\frac{v}{2}\right)^2 = \frac{1}{2} m \frac{v^2}{4} = \frac{1}{8} m v^2

Notice that the initial kinetic energy is

\mathrm{KE} = \frac{1}{2} m v^2.

So we can write:

\mathrm{KE}_{\text{top}} = \frac{1}{8} m v^2 = \frac{1}{4} \left(\frac{1}{2} m v^2\right) = \frac{1}{4} \mathrm{KE}

Thus, the kinetic energy at the highest point is

\frac{\mathrm{KE}}{4}

. Answer: Option C.

Q39

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason

\mathbf{R}

Assertion A: In a central force field, the work done is independent of the path chosen. Reason R: Every force encountered in mechanics does not have an associated potential energy. In the light of the above statements, choose the most appropriate answer from the options given below

A

\mathbf{A}

is false but

\mathbf{R}

is true

B Both

\mathbf{A}

and

\mathbf{R}

are true but

\mathbf{R}

is NOT the correct explanation of

\mathbf{A}

C

\mathbf{A}

is true but

\mathbf{R}

is false

D Both

\mathbf{A}

and

\mathbf{R}

are true and

\mathbf{R}

is the correct explanation of

\mathbf{A}

Correct Answer

Option B

Solution

In a central force field, the work done is independent of the path chosen, this is true because central forces are conservative forces.

Every force encountered in mechanics does not have an associated potential energy this is also true because not all forces are conservative.

Hence, option 2 is correct.

Q40

An object of mass 1000 g experiences a time dependent force

\vec{F}=\left(2 t \hat{i}+3 t^2 \hat{j}\right) N

. The power generated by the force at time

t

is:

A

\left(2 t^2+18 t^3\right) W

B

\left(3 t^3+5 t^5\right) w

C

\left(2 t^3+3 t^5\right) w

D

\left(2 t^2+3 t^3\right) w

Correct Answer

Option C

Solution

Given Information: Force: $\vec{F} = (2t \hat{i} + 3t^2 \hat{j}) \, \text{N}$ Mass of the object, $m = 1000 \, \text{g} = 1 \, \text{kg}$ Determine Acceleration: Using Newton's second law, $\overrightarrow{F} = m \overrightarrow{a}$ .

Thus, the acceleration, $\overrightarrow{a} = 2t \hat{i} + 3t^2 \hat{j}$ .

Velocity Calculation: To find velocity, integrate acceleration with respect to time: $\dfrac{d\overrightarrow{v}}{dt} = 2t \hat{i} + 3t^2 \hat{j}$ Integrate to find $\overrightarrow{v}$ : $\overrightarrow{v} = \int (2t \hat{i} + 3t^2 \hat{j}) \, dt = t^2 \hat{i} + t^3 \hat{j} + \vec{C}$ (Assuming initial velocity is zero, $\vec{C} = 0$ , hence $\overrightarrow{v} = t^2 \hat{i} + t^3 \hat{j}$ ).

Calculate Power: Power is defined as the dot product of force and velocity vectors: $P = \overrightarrow{F} \cdot \overrightarrow{v} = (2t \hat{i} + 3t^2 \hat{j}) \cdot (t^2 \hat{i} + t^3 \hat{j})$ Computing the dot product gives: $P = (2t \cdot t^2) + (3t^2 \cdot t^3) = 2t^3 + 3t^5$ The power generated at time $t$ is $P = (2t^3 + 3t^5) \, \text{W}$ .