Work and Energy

Question 1:

Question 2:

No. Work done in lifting the box increases the potential energy of the box. During lifting at every point, the force applied by us on the box in the upward direction is equal to the gravitational force acting on the box in the downward direction. Therefore, there is no change in the velocity of the box. As a result, the kinetic energy of the box will not change.

Question 3:

No, the kinetic energy of the particle at the bottom of the inclined plane does not depend on the angle of inclination. When the particle reaches the ground, all its potential energy, while at the top of the inclined plane, is converted into kinetic energy. As we know that kinetic energy depends only on the height of the particle, it will be the same for different angles of inclination.

No, we do not need any other information to answer this question.

Question 4:

Yes. Let us consider a block A which is resting on another block B. Block B is resting on a smooth horizontal surface. Let the coefficient of friction between the blocks be $\mathrm{\mu }$.



When a force F is applied on block B in the forward direction as shown in the above figure, block A moves with block B in the direction of the applied force. The frictional force on block A and the displacement will be in the forward direction. Therefore, work done by the frictional force is positive.  

If we consider the reference frame of block B, then displacement of block A will be zero. Therefore, work done by the frictional force is zero.

Question 5:

Yes. Let us consider a block A which is resting on another block B. Block B is resting on a smooth horizontal surface. Let the coefficient of kinetic friction between the blocks be ${\mathrm{\mu }}_{\mathrm{k}}$.



When a force F is applied on block B in the forward direction as shown in the above figure, block A moves with block B in the direction of the applied force. The friction force on block A and the displacement will be in the forward direction. Therefore, work done by the friction force is positive. In this case, block A will remain in contact with block B. This shows that static friction is doing a nonzero work on an object.

Question 6:

Yes. Let us consider an elevator accelerating upward with a body placed in it. In this case, the normal reaction offered by the floor of the elevator on the body is greater than the weight of the body acting in the downward direction. If a person is observing this from the ground, then, for him, the normal reaction is doing a positive work, as the elevator is moving upward.

Question 7:

Yes. Let us consider an isolated system of two particles falling towards each other under their mutual gravitational force of attraction. Here, the net force on the system is zero, but the velocities of the particles keep on increasing. Also, the kinetic energy of the system is increased without applying any external force on it.

Question 8:

In an non-inertial frame, pseudo force also comes into account. As we know that pseudo force does not exist, work-energy theorem is not valid in non-inertial frames.

Question 9:

(i) No. As the surface is smooth and the friction is zero, work done by the force will only depend on the force and the displacement.

(ii) No, because gravitational force is a conservative force and work done by a conservative force will depend only on the force and the displacement.

Question 10:

No, both are correct. We measure potential energy from a reference level chosen by the observer. Therefore, in this case, both observers are measuring the potential energy from different reference levels.

Question 11:

Yes, one of them is necessarily wrong. We measure potential energy from a reference level chosen by the observer. However, the change in potential energy of a body does not depend on the level of reference.

Question 12:

(i) During compression, the work done on the ball is positive as the direction of the force applied by the fingers is along the compression of the ball.

(ii) During expansion, the work done is negative as expansion takes place against the force applied by the fingers on the ball.  

Question 13:

(a) Work by the winning team on the losing team is positive, as the displacement of the losing team is along the force applied by the winning team.
(b) Work by the losing team on the winning team is negative, as the displacement of the winning team is opposite to the force applied by losing team.
(c) Work by the ground on the winning team is positive.
(d) Work by the ground on the losing team is negative.
(e) Total external work on the two teams is positive.

Question 14:

When an apple falls from a tree, its gravitational potential energy decreases as it reaches the ground. After it strikes the ground, its potential energy will remain unchanged.

Question 15:

When a person pushes his bicycle up on an inclined plane, the potential energies of the bicycle and the person increase because moving up on the inclined plane the kinetic energy decreases. and as mechanical energy is sum of kinetic energy and potential energy, and remains constant for a conservative system. Therefore, potential energy must increase in this case.

Question 16:

The magnetic force on a charged particle is always perpendicular to its velocity. Therefore, the work done by the magnetic force on the charged particle is zero. Here, the kinetic energy and speed of the particle remain unaffected, while the velocity changes due to the change in direction of its motion.

Question 17:

(a)
Initial kinetic energy of the ball, $K_i=\frac{1}{2}m{v}^2$
Here, m is the mass of the ball.
The final kinetic of the ball is zero.

(b)
Work done by the kinetic friction is equal to the change in kinetic energy of the ball.
∴ Work done by the kinetic friction = $K_f-K_i=0-\frac{1}{2}m{v}^2$
                                                       = $-\frac{1}{2}m{v}^2$

Question 1:

The relative velocity of the ball w.r.t. the moving frame is given by $v_r=v-v_0$.

(a) Initial kinetic energy of the ball = $\frac{1}{2}m{v_r}^2=\frac{1}{2}m(v-v_0{)}^2$

Also, final kinetic energy of the ball = $\frac{1}{2}m(0-v_0{)}^2=\frac{1}{2}m{v_0}^2$

(b) Work done by the kinetic friction = final kinetic energy $-$ initial kinetic energy
                                                        = $\frac{1}{2}m(v_0{)}^2-\frac{1}{2}m(v-v_0{)}^2$
                                                         = $-\frac{1}{2}m{v}^2+mv{v}_0$

Question 2:

(d) the speed does not depend on the initial direction

As the stone falls under the gravitational force, which is a conservative force, the total energy of the stone remains the same at every point during its motion.

From the conservation of energy, we have:
Initial energy of the stone = final energy of the stone
$\mathrm{i}.\mathrm{e}., (K.E.{)}_i+(P.E.{)}_i=(K.E.{)}_f+(P.E.{)}_f$
$$\begin{gathered} \Rightarrow \frac{1}{2}m{v}^2+mgh=\frac{1}{2}m(v_{max}{)}^2 \\ \Rightarrow v_{max}=\sqrt{v^2+2gh} \end{gathered}$$

From the above expression, we can say that the maximum speed with which stone hits the ground does not depend on the initial direction.

Question 3:

(b) 2E

Let x_A and x_B be the extensions produced in springs A and B, respectively.
Restoring force on spring A, $F=k_{\mathrm{A}}{x}_{\mathrm{A}}$   ...(i)
Restoring force on spring B, $F=k_{\mathrm{B}}{x}_{\mathrm{B}}$   ...(ii)

From (i) and (ii), we get:
$k_{\mathrm{A}}{x}_{\mathrm{A}}=k_{\mathrm{B}}{x}_{\mathrm{B}}$

It is given that k_A = 2k_{B
$\therefore x_{\mathrm{B}}=2x_{\mathrm{A}}$}

Energy stored in spring A:
$E=\frac{1}{2}{k}_{\mathrm{A}}{x_{\mathrm{A}}}^2$   ...(iii)

Energy stored in spring B:
$$\begin{gathered} E\text{'}=\frac{1}{2}{k}_{\mathrm{B}}{x_{\mathrm{B}}}^2=\frac{1}{2}\left(\frac{k_{\mathrm{A}}}{2}\right)(2x_{\mathrm{A}}{)}^2 \\ \therefore E\text{'}=2\times \left(\frac{1}{2}{k}_{\mathrm{A}}{x_{\mathrm{A}}}^2\right)=2E [\mathrm{From} \left(\mathrm{iii}\right)] \end{gathered}$$

Question 4:

(d) $-\frac{1}{4}k{x}^2$

The work done by the spring on both the masses is equal to the negative of the increase in the elastic potential energy of the spring.

The elastic potential energy of the spring is given by $E_p=\frac{1}{2}k{x}^2$.

Work done by the spring on both the masses = $-\frac{1}{2}k{x}^2$
∴ Work done by the spring on each mass = $\frac{1}{2}\left(-\frac{1}{2}k{x}^2\right)=-\frac{1}{4}k{x}^2$

Question 5:

(c) potential energy

The negative of the work done by the conservative internal forces on a system is equal to the changes in potential energy.
i.e. $W=-∆P.E.$

Question 6:

(a) total energy

When work is done by an external forces on a system, the total energy of the system will change.

Question 7:

(a) total energy

The work done by all the forces (external and internal) on a system is equal to the change in the total energy.

Question 8:

(c) Potential energy

The potential energy of a two particle system depends only on the separation between the particles.

Question 9:

(c) mgvt sin²θ

Distance (d) travelled by the elevator in time t = vt
The block is not sliding on the wedge.
Then friction force (f) = mg sin$\theta$

Work done by the friction force on the block in time t is given by
$$\begin{gathered} W=Fd\cos (90-\theta ) \\ \Rightarrow W=mg\sin \theta \times d\times \cos (90-\theta ) \\ \Rightarrow W=mgd{\sin }^2\theta \\ \therefore W=mgvt{\sin }^2\theta \end{gathered}$$

Question 10:

(d) none of these.

The net force on the block is not zero, therefore the block will not be in any given equilibrium.

Question 1:

(c) $\sqrt{3gl}$

Suppose that one end of an extensible string is attached to a mass m, while the other end is fixed. The mass moves with a velocity v in a vertical circle of radius R. At some instant, the string makes an angle θ with the vertical as shown in the figure.



For a complete circle, the minimum velocity at L must be $v_{\mathrm{L}}=\sqrt{5gl}$.

Applying the law of conservation of energy, we have:

Total energy at M = total energy at L
$$\begin{gathered} \mathrm{i}.\mathrm{e}., \frac{1}{2}m{v_{\mathrm{M}}}^2+mgl=\frac{1}{2}m{v_{\mathrm{L}}}^2 \\ \Rightarrow \frac{1}{2}m{v_{\mathrm{M}}}^2=\frac{1}{2}m{v_{\mathrm{L}}}^2-mgl \\ \mathrm{Using} v_L\ge \sqrt{5gl}, \mathrm{we} \mathrm{have}: \\ \frac{1}{2}m{v_{\mathrm{M}}}^2\ge \frac{1}{2}m\left(5gl\right)-mgl \\ \therefore v_{\mathrm{M}}=\sqrt{3gl} \end{gathered}$$

Question 2:

(a) must depend on the speed of projection
(b) must be larger than the speed of projection

Consider that the stone is projected with initial speed v.

As the stone is falls under the gravitational force, which is a conservative force, the total energy of the stone remains the same at every point during its motion.

From the conservation of energy, we have:
Initial energy of the stone = final energy of the stone
$\mathrm{i}.\mathrm{e}., (K.E.{)}_i+(P.E.{)}_i=(K.E.{)}_f+(P.E.{)}_f$
$$\begin{gathered} \Rightarrow \frac{1}{2}m{v}^2+mgh=\frac{1}{2}m(v_{max}{)}^2 \\ \Rightarrow v_{max}=\sqrt{v^2+2gh} \end{gathered}$$

From the above expression, we can say that the maximum speed with which the stone hits the ground depends on the speed of projection and greater than it.

Question 3:

(a) always

According to the work-energy theorem, the total work done on a particle is equal to the change in kinetic energy of the particle.

Question 4:

(c) its kinetic energy is constant
(d) it moves in a circular path.

When the force on a particle is always perpendicular to its velocity, the work done by the force on the particle is zero, as the angle between the force and velocity is 90$°$. So, kinetic energy of the particle will remain constant. The force acting perpendicular to the velocity of the particle provides centripetal acceleration that causes the particle to move in a circular path.

Question 5:

(d) acceleration of the block

Acceleration of the block will be the same to both the observers. The respective kinetic energies of the observers are different, because the block appears to be moving with different velocities to both the observers. Work done by the friction and the total work done on the block are also different to the observers. 

Question 6:

(a) the path taken by the suitcase
(b) the time taken by you in doing so
(d) your weight

Work done by us on the suitcase is equal to the change in potential energy of the suitcase.
i.e., W = mgh
Here, mg is the weight of the suitcase and h is height of the table.

Hence, work done by the conservative (gravitational) force does not depend on the path.

Question 7:

(a) the force is always perpendicular to its velocity
(c) the object is stationary but the point of application of the force moves on the object
(d) the object moves in such a way that the point of application of the force remains fixed.

No work is done by a force on an object if the force is always perpendicular to its velocity. Acceleration does not always provide the direction of motion, so we cannot say that no work is done by a force on an object if it is always perpendicular to the acceleration. Work done is zero when the displacement is zero.
In a circular motion, force provides the centripetal acceleration. The angle between this force and the displacement is 90$°$, so work done by the force on an object is zero.

Question 8:

(a) The string becomes slack when the particle reaches its highest point.
(d) The particle again passes through the initial position.

The string becomes slack when the particle reaches its highest point. This is because at the highest point, the tension in the string is minimum. At this point, potential energy of the particle is maximum, while its kinetic energy is minimum. From the law of conservation of energy, we can see that the particle again passes through the initial position where its potential energy is minimum and its kinetic energy is maximum.

Question 9:

(b) The resultant force on the particle must be at an angle less than 90° with the velocity all the time.
(d) The magnitude of its linear momentum is increasing continuously.

Kinetic energy of a particle is directly proportional to the square of its velocity. The resultant force on the particle must be at an angle less than 90° with the velocity all the time so that the velocity or kinetic energy of the particle keeps on increasing.
The kinetic energy is also directly proportional to the square of its momentum, therefore it continuously increases with the increase in momentum of the particle. 

Question 10:

(a) at spring was initially compressed by a distance x and was finally in its natural length
(b) it was initially stretched by a distance x and and finally was in its natural length

For an elastic spring, the work done is equal to the negative of the change in its potential energy.

When the spring was initially compressed or stretched by a distance x, its potential energy is given by
${\left(P.E.\right)}_i=\frac{1}{2}k{x}^2$.

When it finally comes to its natural length, its potential energy is given by
${\left(P.E.\right)}_f=0$.

∴ Work done = $-\left[{\left(P.E.\right)}_f-{\left(P.E.\right)}_i\right]=-\left[0-\frac{1}{2}k{x}^2\right]=\frac{1}{2}k{x}^2$

Question 1:

(b) The tension in the string is F.

Tension in the string is equal to F, as tension on both sides of a frictionless and massless pulley is the same.  
i.e., T – Mg = Ma
$\Rightarrow$T = Mg + Ma
So, the tension in the string cannot be equal to Mg.
The change in kinetic energy of the block is equal to the work done by gravity.
Hence, the work done by gravity is 20 J in 1 s, while the the work done by the tension force is zero.

Question 2:

Total mass of the system (cyclist and bike), $M=m_c+m_b=90 \mathrm{kg}$

Initial velocity of the system, $u=6.0 \mathrm{km}/\mathrm{h}=1.666 \mathrm{m}/\sec$

Final velocity of the system, $\nu =12 \mathrm{km}/\mathrm{h}=3.333 \mathrm{m}/\sec$

From work-energy theorem, we have:
$$\begin{gathered} \mathrm{Increase} \mathrm{in} \mathrm{K}.\mathrm{E}.=\frac{1}{2}M{\nu }^2-\frac{1}{2}m{u}^2 \\ =\frac{1}{2}90\times {\left(3.333\right)}^2-\frac{1}{2}\times 90\times {\left(1.66\right)}^2 \\ =499.4-124.6 \\ =374.8=375 \mathrm{J} \end{gathered}$$

Question 3:

$$\begin{gathered} \mathrm{Mass} \mathrm{of} \mathrm{the} \mathrm{block}, M_b=2 \mathrm{kg} \\ \mathrm{Initial} \mathrm{speed} \mathrm{of} \mathrm{the} \mathrm{block}, u=10 \mathrm{m}/s \\ \mathrm{Also}, a=3 \mathrm{m}/s^2 \mathrm{and} t=5 \mathrm{s} \\ \mathrm{Using} \mathrm{the} \mathrm{equation} \mathrm{of} \mathrm{the} \mathrm{motion}, \mathrm{we} \mathrm{have}: \\ \nu =u+at \\ =10+3\times 5=25 \mathrm{m}/s \\ \therefore \mathrm{Final} \mathrm{K}.\mathrm{E}.=\frac{1}{2}m{\nu }^2 \\ =\frac{1}{2}\times 2\times 625=625 \mathrm{J} \end{gathered}$$

Question 4:

Resisting force acting on the box, $F=100 \mathrm{N}$
Displacement of the box, S = 4 m
Also,$\theta =180°$

∴ Work done by the resisting force, $\mathrm{W}=\underset{\mathrm{F}}{\to }·\underset{\mathrm{S}}{\to }=100\times 4\times \cos 180°=-400 \mathrm{J}$

Question 5:

$$\begin{gathered} \mathrm{Mass} \mathrm{of} \mathrm{the} \mathrm{block}, M=5 \mathrm{kg} \\ \mathrm{Angle} \mathrm{of} \mathrm{inclination}, \theta =30° \end{gathered}$$
Gravitational force acting on the block, $\mathrm{F}=mg$
Work done by the force of gravity depends only on the height of the object, not on the path length covered by the object.
   

$$\begin{gathered} \mathrm{Height} \mathrm{of} \mathrm{the} \mathrm{object}, h=10\times \sin 30° \\ =10\times \frac{1}{2}=5 \mathrm{m} \end{gathered}$$

$$\begin{gathered} \therefore \mathrm{Work} \mathrm{done} \mathrm{by} \mathrm{the} \mathrm{force} \mathrm{of} \mathrm{gravity}, w=mgh \\ =5\times 9.8\times 5=245 \mathrm{J} \end{gathered}$$

Question 6:

Given:
$F=2.50 \mathrm{N}, S=2.5 \mathrm{m} \mathrm{and} m=15 g=0.015 \mathrm{kg}$

Work done by the force,
$$\begin{gathered} W=F\mathit{·}S \cos 0° \left(\mathrm{acting} \mathrm{along} \mathrm{the} \mathrm{same} \mathrm{line}\right) \\ =2.5\times 2.5=6.25 \mathrm{J} \end{gathered}$$

Acceleration of the particle is,

$$\begin{gathered} a=\frac{F}{m}=\frac{2.5}{0.015} \\ =\frac{500}{3} \mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Applying the work-energy principle for finding the final velocity of the particle,
$$\begin{gathered} \frac{1}{2}m{v}^2-0=6.25 \\ \Rightarrow \nu =\sqrt{\frac{6.25\times 2}{0.15}}=28.86 \mathrm{m}/s \end{gathered}$$
So, time taken by the particle to cover 2.5 m distance,
$$\begin{gathered} t=\frac{\nu -u}{\alpha }=\frac{\left(28.86\right)\times 3}{500} \\ \therefore \mathrm{Average} \mathrm{power}=\frac{W}{t} \\ =\frac{6.25\times 500}{\left(28.86\right)\times 3}=36.1 \mathrm{W} \end{gathered}$$

Question 7:

Initial position vector,
$\overrightarrow{r_1}=2\overrightarrow{i}+3\overrightarrow{j}$
Final position vector,
${\overrightarrow{r}}_2=3\overrightarrow{i}+2\overrightarrow{j}$
So, displacement vector,
$$\begin{gathered} \overrightarrow{r}={\overrightarrow{r}}_2-{\overrightarrow{r}}_1 \\ =\left(3\overrightarrow{i}+2\overrightarrow{j}\right)-\left(2\overrightarrow{i}+\overrightarrow{j}\right) \\ =\overrightarrow{i}-\overrightarrow{j} \\ \mathrm{Force} \mathrm{acting} \mathrm{on} \mathrm{the} \mathrm{particle}, \overrightarrow{\mathit{F}}=5\overrightarrow{i}+5\overrightarrow{j} \\ \mathrm{So}, \mathrm{work} \mathrm{done}=\overrightarrow{\mathit{F}}\mathit{·}\overrightarrow{\mathit{S}}=5\times 1+5 \left(-1\right)=0 \end{gathered}$$

Question 8:

Given:
$$\begin{gathered} \mathrm{Mass} \mathrm{of} \mathrm{the} \mathrm{block}, m=2 \mathrm{kg} \\ \mathrm{Distance} \mathrm{coverd} \mathrm{by} \mathrm{the} \mathrm{man}, s=40 \mathrm{m} \\ \mathrm{Acceleration} \mathrm{of} \mathrm{the} \mathrm{man}, \mathrm{a}=0.5 \mathrm{m}/s^2 \end{gathered}$$
So, force applied by the man on the box,
$$\begin{gathered} F=ma \\ =2\times \left(0.5\right) \\ =1 \mathrm{N} \\ \mathrm{Work} \mathrm{done} \mathrm{by} \mathrm{the} \mathrm{man} \mathrm{on} \mathrm{the} \mathrm{block}, W=F\mathit{·}S \\ =1\times 40 \\ =40 \mathrm{J} \end{gathered}$$

Question 9:

Given that force is a function of displacement, i.e. $\mathrm{F}=a+bx$,
where a and b are constants.
So, work done by this force during the displacement x = 0 to x = d,
$$\begin{gathered} \mathrm{W}=\underset{0}{\overset{d}{\int }}\mathrm{F} dx \\ W=\underset{0}{\overset{d}{\int }} \left(a+bx\right) dx \\ W={\left[ax+\frac{b{x}^2}{2}\right]}_0^d \\ W=ad+\frac{b{d}^2}{2} \\ \Rightarrow W=\left(a+\frac{bd}{2}\right)d \end{gathered}$$

Question 10:

$$\begin{gathered} \mathrm{Given}: \\ m=250 g, \mathrm{\theta }=37°, d=1 \mathrm{m} \end{gathered}$$

Here, R is the normal reaction of the block.



As the block is moving with uniform speed,
$f=mg\sin 37°$

So, work done against the force of friction,
$$\begin{gathered} W=fd\cos 0° \\ W=(mg\sin 37°)\times d \\ W=(0.25\times 9.8\times \sin 37°)\times 1.0 \\ W=1.5 \mathrm{J} \end{gathered}$$

Question 11:

(a)
$a=\frac{\mathrm{F}}{2 \left(\mathrm{M}+m\right)} \left(\mathrm{given}\right)$

The free-body diagrams of both the blocks are shown below:




For the block of mass m,

$$\begin{gathered} ma={\mu }_1{\mathrm{R}}_1 \mathrm{and} {\mathrm{R}}_1=mg \\ \Rightarrow {\mu }_1=\frac{ma}{{\mathrm{R}}_1} \\ =\frac{F}{2 \left(\mathrm{M}+m\right) g} \end{gathered}$$

(b)
Frictional force acting on the smaller block,
$$\begin{gathered} f_1={\mu }_1{\mathrm{R}}_1=\frac{\mathrm{F}}{2 \left(\mathrm{M}+m\right) g}\times mg \\ =\frac{m\mathrm{F}}{2 (\mathrm{M}+m)} \end{gathered}$$

(c) Work done, w = f₁s [where s = d]
$$\begin{gathered} =\frac{m\mathrm{F}}{2 \left(\mathrm{M}+m\right)}\times d \\ =\frac{m\mathrm{F}d}{2 \left(\mathrm{M}+m\right)} \end{gathered}$$

Question 12:

Given:
$\mathrm{Weight}=2000 \mathrm{N}, \mathrm{s}=20 \mathrm{m}, \mu =0.2$

The free-body diagram for the box is shown below:




(a) From the figure,
  $$\begin{gathered} R+P \sin \mathrm{\theta }-2000=0 ... \left(i\right) \\ P \cos \mathrm{\theta }-0.2 R=0 ... \left(ii\right) \end{gathered}$$

From (i) and (ii),
$$\begin{gathered} P \cos \mathrm{\theta }-0.2 \left(2000-P \sin \mathrm{\theta }\right)=0 \\ P \left(\cos \mathrm{\theta }+0.2 \sin \mathrm{\theta }\right)=400 \\ P=\frac{400}{\cos \mathrm{\theta }+0.2 \sin \mathrm{\theta }} ... \left(iii\right) \end{gathered}$$

So, work done by the person,

$$\begin{gathered} W\mathit{=}PS \cos \mathrm{\theta } \\ =\frac{8000 \cos \mathrm{\theta }}{\cos \mathrm{\theta }+0.2 \sin \mathrm{\theta }} \\ =\frac{8000}{1+0.2 \tan \mathrm{\theta }} \\ =\frac{40000}{5+\tan \mathrm{\theta }} ...\left(iv\right) \end{gathered}$$

(b) For minimum magnitude of force from equation (iii),
$$\begin{gathered} \frac{d}{dk} \left(\cos \mathrm{\theta }+0.2 \sin \mathrm{\theta }\right)=0 \\ \Rightarrow \tan \mathrm{\theta }=0.2 \end{gathered}$$
Putting the value in equation (iv),
$$\begin{gathered} W=\frac{40000}{\left(5+\tan \mathrm{\theta }\right)} \\ =\frac{40000}{5+0.2}\approx 7692 \mathrm{J} \end{gathered}$$

Question 13:

Given:
$$\begin{gathered} Weight, mg=100 \mathrm{N} \\ \mathrm{\theta }=37° \mathrm{and} s=2 \mathrm{m} \\ \mathrm{Force}, \mathrm{F}=mg \sin 37° \\ =100\times \frac{3}{5}=60 \mathrm{N} \end{gathered}$$
So, work done when the force is parallel to incline,
$$\begin{gathered} W=FS \cos \mathrm{\theta } \\ =60\times 2\times \cos 0° =120 \mathrm{J} \end{gathered}$$




$$\begin{gathered} \mathrm{In} \Delta \mathrm{ABC}, \mathrm{AB}=2 \mathrm{m} \\ \mathrm{AC}=h \\ =s\times \sin 37°=2.0\times \sin 37° \\ =1.2 \mathrm{m} \end{gathered}$$

∴ Work done when the force is in horizontal direction,

$$\begin{gathered} W\text{'}=mgh \\ =100\times 1.2=120 \mathrm{J} \end{gathered}$$

Question 14:

$$\begin{gathered} \text{Given}: \\ \mathrm{Mass} \mathrm{of} \mathrm{the} \mathrm{car}, m=500 \mathrm{kg} \\ \mathrm{Distance} \mathrm{covered} \mathrm{by} \mathrm{the} \mathrm{car}, s=25 \mathrm{m} \\ \mathrm{Initial} \mathrm{speed} \mathrm{of} \mathrm{the} \mathrm{car}, u=72 \mathrm{km}/\mathrm{h}=20 m/s \\ \mathrm{Final} \mathrm{speed} \mathrm{of} \mathrm{the} \mathrm{car}, \nu =0 m/s \end{gathered}$$
Retardation of the car,
$$\begin{gathered} a=\frac{\left({\nu }^2-u^2\right)}{2s} \\ \Rightarrow a=-\frac{400}{50}=-8 m/s^2 \\ \mathrm{Frictional} \mathrm{force}, F=ma=500\times 8=4000 \mathrm{N} \end{gathered}$$

Question 15:

$$\begin{gathered} \text{Given}: \\ \mathrm{Mass} \mathrm{of} \mathrm{the} \mathrm{car}, m=500 \mathrm{kg} \\ \mathrm{Initial} \mathrm{velocity} \mathrm{of} \mathrm{the} \mathrm{car}, u=0 \\ \mathrm{Final} \mathrm{velocity} \mathrm{of} \mathrm{the} \mathrm{car}, \nu =72 \mathrm{km}/\mathrm{h}=20 \mathrm{m}/s \\ a=\frac{\left({\nu }^2-u^2\right)}{2s} \\ a=\frac{400}{50}=8 \mathrm{m}/s^2 \end{gathered}$$

Force needed to accelerate the car,
$\mathrm{F}=ma=500\times 8=400\mathrm{N}$

Question 16:

Given,
$$\begin{gathered} \nu =a\sqrt{x} \left(\mathrm{uniformly} \mathrm{accelerated} \mathrm{motion}\right) \\ \text{D}\mathrm{isplacement}, s=d-0=d \\ \text{P}\mathrm{utting} x=0, \text{we get} {\nu }_1=0 \\ \text{P}\mathrm{utting} x=d, \text{we get } {\nu }_2=a\sqrt{d} \\ \alpha =\frac{{\nu }_2^2-{\nu }_1^2}{2s}=\frac{a^{2}d}{2d}=\frac{a^2}{2} \end{gathered}$$
$$\begin{gathered} \mathrm{Force}, F=m\alpha =\frac{m{a}^2}{2} \\ \mathrm{Work} \mathrm{done}, \mathrm{W}=Fs \cos \mathrm{\theta } \\ =\frac{m{a}^2}{2}\times d=\frac{m{a}^{2}d}{2} \end{gathered}$$

Question 17:

(a)
Given:
$$\begin{gathered} \mathrm{Mass} \mathrm{of} \mathrm{the} \mathrm{block}, m=2 \mathrm{kg} \\ \mathrm{\theta }=37° \\ \mathrm{Force} \mathrm{on} \mathrm{the} \mathrm{block}, F=20 \mathrm{N} \end{gathered}$$



From the above figure,

$$\begin{gathered} F=\left(2g \sin \mathrm{\theta }\right)+ma \\ \Rightarrow a=\frac{20-20 \sin \mathrm{\theta }}{2}=4 m/s^2 \\ s=ut+\frac{1}{2}a{t}^2=2 \mathrm{m} \\ \mathrm{So}, \mathrm{work} \mathrm{done} \\ W=FS=20\times 2=40 \mathrm{J} \end{gathered}$$



(b) If $W=40 \mathrm{J}$
$$\begin{gathered} S\mathit{=}\frac{\mathit{W}}{\mathit{F}}=\frac{40}{20}=2 \mathrm{m} \\ h=2 \sin 37°=1.2 \mathrm{m} \end{gathered}$$


So, work done
$$\begin{gathered} \mathrm{W}=-mgh \\ =-20\times 1.2=-24 \mathrm{J} \end{gathered}$$

(c)
$$\begin{gathered} \nu =u+at \\ =4\times 1=4 \mathrm{m}/\sec \\ \mathrm{So}, \mathrm{K}.\mathrm{E}.=\frac{1}{2} m{v}^2 \\ =\frac{1}{2}\times 2\times 16=16 \mathrm{J} \end{gathered}$$

Question 18:

Given:
$$\begin{gathered} \mathrm{Mass}, m=2 \mathrm{kg} \\ \mathrm{Inclination}, \mathrm{\theta }=37° \\ \mathrm{Force} \mathrm{applied}, \mathrm{F}=20 \mathrm{N} \\ \mathrm{Acceleration} \mathrm{of} \mathrm{the} \mathrm{block}, a=10 \mathrm{m}/s^2 \end{gathered}$$
(a) t = 1 sec

So, $s=ut+\frac{1}{2}a{t}^2=5 \mathrm{m}$





Work done by the applied force,
$W\mathit{=}Fs \cos \theta °=20\times 5=100 \mathrm{J}$

(b) $\mathrm{AB} \left(\mathrm{h}\right)=5 \sin 37°=3 \mathrm{m}$
So, work done by weight,
$$\begin{gathered} \mathrm{W}=mgh \\ 2\times 10\times 3=60 \mathrm{J} \end{gathered}$$
So, frictional force,
$f=mg \sin \mathrm{\theta }$
Work done by the friction forces,
$$\begin{gathered} \mathrm{W}=fs \cos 0°=\left(mg \sin \mathrm{\theta }\right) s \\ =20\times 0.60\times 5=-60 \mathrm{J} \end{gathered}$$

Question 19:

$$\begin{gathered} \mathrm{Given}: \\ \mathrm{Mass} \mathrm{of} \mathrm{the} \mathrm{block}, m=250 \mathrm{gm}=0.250 \mathrm{kg} \\ \mathrm{Initial} \mathrm{speed} \mathrm{of} \mathrm{the} \mathrm{block}, u=40 \mathrm{cm}/s=0.4 \mathrm{m}/s \\ \mathrm{Final} \mathrm{speed} \mathrm{of} \mathrm{the} \mathrm{block}, \nu =0 \\ \mathrm{Coefficient} \mathrm{of} \mathrm{friction}, \mu =0.1 \end{gathered}$$

Force in the forward direction is equal to the friction force.
$$\begin{gathered} \mathrm{Here}, \mu \mathrm{R}=ma \\ \left(\mathrm{where} a \mathrm{is} \mathrm{deceleration}\right) \\ a=\left(\frac{\mu \mathrm{R}}{m}\right)=\left(\frac{\mu mg}{m}\right)=\mu g \\ =0.1\times 9.8=0.98 m/s^2 \\ s=\frac{{\nu }^2-u^2}{2a}=0.082 \mathrm{m} \\ = 8.2 \mathrm{cm} \end{gathered}$$

Again, work done against friction,

$$\begin{gathered} \mathrm{W}=-\mu Rs \cos \mathrm{\theta } \\ =-1\times 2.5\times 0.082\times 1 \\ =-0.02 \mathrm{J} \\ \Rightarrow \mathrm{W}=-0.02 \mathrm{J} \end{gathered}$$