Work, Energy and Power

To find the speed of the mass just before it touches the spring, we begin by determining the spring constant, $ k $, using Hooke's Law, where the force $ F $ is proportional to the compression $ x $:

$ F = kx $

Given that a force of 100 N compresses the spring by 1 m, the spring constant $ k $ is calculated as:

$ k = \frac{F}{x} = \frac{100 \, \text{N}}{1 \, \text{m}} = 100 \, \text{N/m} $

Next, we apply the principle of energy conservation. The initial potential and kinetic energy will equal the final potential energy stored in the compressed spring. The relevant energy conservation equation is:

$ \frac{1}{2} mv^2 + mgh = \frac{1}{2} kx_{\text{max}}^2 $

where:

$ m $ is the mass of the block (10 kg),

$ g $ is the acceleration due to gravity (10 m/s$^2$),

$ h $ is the vertical height change, which is the same as the compression in this scenario ($x_{\text{max}} = 2 \, \text{m}$),

$ v $ is the speed just before touching the spring,

$ x_{\text{max}} $ is the maximum compression of the spring (2 m).

Rearranging the equation for $ v $ gives:

$ v = \sqrt{\frac{k x_{\text{max}}^2}{m} - 2gh} $

Substituting the known values:

$ v = \sqrt{\frac{(100 \, \text{N/m})(2 \, \text{m})^2}{10 \, \text{kg}} - 2(10 \, \text{m/s}^2)\left(\frac{2}{2}\right)} $

This simplifies to:

$ v = \sqrt{\frac{(100)(4)}{10} - 40} $

$ v = \sqrt{40 - 20} = \sqrt{20} \, \text{m/s} $

Thus, the speed of the mass just before it touches the spring is $ \sqrt{20} \, \text{m/s} $.

Given,

Mass pulled by engine $m=5000 \mathrm{~kg}$

Velocity, $v=5 \mathrm{~m} / \mathrm{s}$

Inclination $=1 / 50$

Let $\theta$ be the angle of inclination

$ \sin \theta=1 / 50 $

Force acting on mass on inclined plane $F_{\text {net }}=m g \sin \theta+$ frictional force Frictional force $=0$ [Smooth surface]

$ \begin{aligned} F_{\mathrm{net}} & =F=5000 \times 10 \times \frac{1}{50} \quad\left[g=10 \mathrm{~m} / \mathrm{s}^2\right] \\ F & =1000 \mathrm{~N} \end{aligned} $

Now power is given by

$ \begin{aligned} P & =F \cdot v=1000 \times 5 \\ & =5000 \mathrm{~W}=5 \mathrm{~kW} \end{aligned} $

Let us assume that the displacement of the body is directly proportional to $t^n$ is

$\begin{aligned} & \text { } \\ & \text { } \quad \begin{aligned} s & =K t^n \\and\,\,\,\,\, v & =\frac{d S}{d t}=K n t^{n-1} \\and\,\,\,\, a & =\frac{d v}{d t}=K(n-1) n t^{n-2}\end{aligned} \\ & \text { Force }=m a=m K n(n-1) t^{n-2} \\ & \text { Power, } \quad \begin{aligned} P & =F v \\ & =m K n(n-1) t^{n-2} \times K n t^{n-1} \\ & =m K^2 n^2(n-1) t^{2 n-3}\end{aligned}\end{aligned}$

As power is constant and independent of time, hence $2 n-3=0$

$ \begin{aligned} or\,\,\,& n=\frac{3}{2} \\ & s \propto t^{3 / 2} \end{aligned} $

Given, mass $m=3 \mathrm{~kg}$

displacement $(s)=\frac{t^3}{3} \mathrm{~m}$

Velocity, $v=\frac{d s}{d t}=\frac{3 t^2}{3} \Rightarrow t^2 \mathrm{~m} / \mathrm{s}$

Acceleration, $(a)=\frac{d v}{d t}=2 t \mathrm{~m} / \mathrm{s}^2$

Now, work done by the force

$ \begin{aligned} W & =\int_0^2 F \cdot d s=\int_0^2 m a \cdot d s \\ & =\int_0^2 3 \times 2 t \times t^2 d t \\ & =\int_0^2 6 t^3 d t=\frac{3}{2}\left[t^4\right]_0^2 \\ & =24 \mathrm{~J} \end{aligned} $

When a person climbs stairs, their gravitational potential energy increases. But where does this energy come from? The source is the work done by internal forces within the person’s body.

Let's break it down:

Consider a person starting at the ground level, where the initial height $ h = 0 $.

As the person climbs and reaches a height $ h $, their potential energy increases by:

$ mgh $

where $ m $ is the mass of the person, $ g $ is the acceleration due to gravity, and $ h $ is the height climbed.

The change in potential energy is mathematically expressed as:

$ \text{PE}_f - \text{PE}_i = \int dW $

where $ dW $ represents the work done.

In this scenario, the increase in gravitational potential energy results from the work done by the internal forces within the person’s body, such as the muscles exerting force to climb the stairs.

Given two bodies with masses $ m_1 = 1 \, \text{g} $ and $ m_2 = 4 \, \text{g} $, both are moving with equal kinetic energies. To find the ratio of their linear momenta, we can use the formula for linear momentum, which is related to kinetic energy as follows:

$ p = \sqrt{2m(\text{KE})} $

For the first body:

$ p_1 = \sqrt{2 \cdot 1 \cdot \text{KE}} $

For the second body:

$ p_2 = \sqrt{2 \cdot 4 \cdot \text{KE}} $

To find the ratio of the magnitudes of their linear momenta, divide the expressions of $ p_1 $ and $ p_2 $:

$ \frac{p_1}{p_2} = \frac{\sqrt{2 \cdot \text{KE}}}{\sqrt{8 \cdot \text{KE}}} $

Since the kinetic energies are equal, they cancel out in the equation:

$ \frac{p_1}{p_2} = \sqrt{\frac{1}{4}} = \frac{1}{2} = 1:2 $

Thus, the ratio of the magnitudes of their linear momenta is $ 1:2 $.

Given that the displacement $ s $ of a body of mass 3 kg under the action of a force is defined by the equation $ s = \frac{t^3}{3} $, where $ s $ is in meters and $ t $ is in seconds, we need to find the work done by the force in the first two seconds.

Here is how we can calculate it:

Displacement Equation:

$ s = \frac{t^3}{3} $

Calculate Velocity:

Velocity $ v $ is the derivative of displacement with respect to time:

$ v = \frac{d s}{d t} = t^2 $

Calculate Acceleration:

Acceleration $ a $ is the derivative of velocity with respect to time:

$ a = \frac{d v}{d t} = 2t $

Force Equation:

Force $ F $ is mass times acceleration:

$ F = m \cdot a = 3 \times (2t) = 6t $

Work Done Calculation:

Work $ W $ is the integral of $ F $ with respect to displacement $ ds $:

$ W = \int F \cdot ds = \int_{0}^{2} 6t \cdot t^2 \, dt = \int_{0}^{2} 6t^3 \, dt $

Evaluate the Integral:

$ \begin{aligned} W &= 6 \int_{0}^{2} t^3 \, dt \\ &= 6 \left[\frac{t^4}{4}\right]_{0}^{2} \\ &= 6 \left[\frac{16}{4} - 0\right] \\ &= 6 \times 4 \\ &= 24 \, \text{Joules} \end{aligned} $

Hence, the work done by the force in the first two seconds is 24 Joules.

To find the force acting on a particle in a force field given by a potential energy function, we can use the relationship between force and potential energy. The force is the negative gradient of the potential energy function:

$\mathbf{F} = -\nabla u(x, y)$

Given the potential energy function:

$u(x, y) = \cos(x + y)$

The gradient operator $\nabla$ in two dimensions is given by:

$\nabla = \left(\frac{\partial}{\partial x}, \frac{\partial}{\partial y}\right)$

To find the force, we first need to compute the partial derivatives of $u(x, y)$ with respect to $x$ and $y$.

Calculate the partial derivative of $u(x, y)$ with respect to $x$:

$\frac{\partial u}{\partial x} = \frac{\partial}{\partial x}\cos(x + y) = -\sin(x + y)$

Now, calculate the partial derivative of $u(x, y)$ with respect to $y$:

$\frac{\partial u}{\partial y} = \frac{\partial}{\partial y}\cos(x + y) = -\sin(x + y)$

So the gradient of $u(x, y)$ is:

$\nabla u(x, y) = \left(-\sin(x + y), -\sin(x + y)\right)$

The force $\mathbf{F}$ is the negative of this gradient:

$\mathbf{F} = -\nabla u(x, y) = \left(\sin(x + y), \sin(x + y)\right)$

We need to evaluate this at the coordinates $(0, \pi / 6)$:

$ (x, y) = (0, \pi / 6) $

Substitute these values into the sine function:

$ \sin(0 + \pi / 6) = \sin(\pi / 6) = \frac{1}{2} $

Thus, the force $\mathbf{F}$ at $(0, \pi / 6)$ is:

$\mathbf{F} = \left(\frac{1}{2}, \frac{1}{2}\right)$

In vector notation, this can be written as:

$ \mathbf{F} = \frac{1}{2} \hat{\mathbf{i}} + \frac{1}{2} \hat{\mathbf{j}} $

This matches with Option B:

$\frac{1}{2}(\hat{\mathbf{i}}+\hat{\mathbf{j}})$

Net force on boat,

$ F_{\mathrm{net}}=F_{\mathrm{boat}}-F_{\mathrm{drag}} $

Average power of boat

$ =\frac{1}{2} F_{\mathrm{net}} \times v $

$ \begin{aligned} & \text { So, } \quad \frac{1}{2}\left(F_{\text {boat }}-F_{\text {drag }}\right) \times v=P \\ & \Rightarrow \quad \frac{1}{2}\left(m \times a_{\text {boat }}-F_{\text {drag }}\right) \times v=P\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \end{aligned} $

Given, $m=1000 \mathrm{~kg}$

$ \begin{aligned} a_{\text {boat }} & =\frac{20-0}{5}=4 \mathrm{~m} / \mathrm{s}^2 \\ v & =20 \mathrm{~m} / \mathrm{s} \\ \Rightarrow \quad P & =45000 \mathrm{~W} \end{aligned} $

Substituting given data, in Eq. (i), We get

$ \frac{1}{2}\left(1000 \times 4-F_{\mathrm{drag}}\right) \times 20=45000 $

$ \begin{array}{rlr} \Rightarrow 4000-F_{\text {drag }} =4500 \\ \text { or } F_{\text {drag }} =500 \mathrm{~N} \end{array} $

(negative sign because $F_{\text {drag }}$ opposes velocity)

$ \begin{aligned} \text { Work done by pump } & =m g h \\ & =V \cdot \rho \cdot g \cdot h \end{aligned} $

$ \begin{aligned} & \begin{array}{l} V=\text { volume }, \rho=\text { density, } h=\text { height } \\ \text { Power output of pump } \\ \quad=\frac{\text { Work done by pump }}{\text { Time taken }} \\ \quad=\frac{V \cdot \rho \cdot g \cdot h}{t} \end{array} \end{aligned} $

Efficiency of pump

$ \begin{aligned} & =\frac{\text { Power output }}{\text { Electrical power consumed }} \times 100 \\ \eta & =\frac{V \cdot \rho \cdot g \cdot h \times 100}{P_{\text {electric }} \times t} \end{aligned} $

$ \begin{aligned}Here, V & =36 \mathrm{~m}^3, \rho=1000 \mathrm{~kg} / \mathrm{m}^3 \\ h & =50 \mathrm{~m} \\ P & =40 \times 1000 \mathrm{~W}, t=30 \mathrm{~min} \end{aligned} $

So, efficiency

$ \begin{aligned} \eta & =\frac{36 \times 1000 \times 10 \times 50 \times 100}{40 \times 1000 \times 30 \times 60} \\ \Rightarrow \quad \eta & =25 \% \end{aligned} $

Kinetic energy $K$ and displacement $x$ are related as $K=F . x$ (work-energy theorem)

So, slope of $K \cdot x$ curve is $\frac{d K}{d x}=\frac{d}{d x} F x=F$ (let force is constant)

∴ Slope of $K-x$ curve $=$ Force $=$ Mass × Acceleration

or $\quad \frac{d K}{d x} \propto a$ (statement I is correct)

Statement II is incorrect because ball rises again to same height as it was dropped, therefore it is the case of perfectly elastic collision and hence, no loss of energy occurs.

Also, for a body starting from rest, $\mathrm{KE}=$ Work done

$ \begin{array}{rlrl} \Rightarrow & & \frac{1}{2} m v^2 & =F x=m \cdot a \cdot x \\ \Rightarrow & v & v & =\sqrt{2 a x} \text { which is mass independent. } \end{array} $

∴ Statement III is incorrect.

                        $ \begin{array}{ll} P=m a v & (\because F=m a) \\ P=m \frac{d v}{d t} v & \end{array} $

$ \begin{aligned} \Rightarrow & & v d v & =\frac{P}{m} d t \\ \Rightarrow & & \int v d v & =\frac{P}{m} \int d t \\ \text { or } & & \frac{v^2}{2} & =\frac{P t}{m} \\ \Rightarrow & & v & =\left(\frac{2 P}{m}\right)^{1 / 2} \cdot t^{1 / 2} \\ \text { or } & & v & \propto t^{1 / 2} \end{aligned} $

Work done,

$ \begin{aligned} W & =\int P \cdot d t=\int F \cdot v \cdot d t \\ & =\int \frac{m d v}{d t} \cdot v \cdot d t \\ & =\int m \cdot c \cdot \alpha \cdot x^{\alpha-1} \cdot \frac{d x}{d t} \cdot v \cdot d t \\ & =\int m c^2 x^{2 \alpha-1} \cdot \alpha \cdot d x \end{aligned} $

Here, $m=1 \mathrm{~kg}, c=1$ and $x_i=0, x_C=4 \mathrm{~m}$

$ \begin{aligned}So, W & =128=-\int_0^4 \alpha x^{2 \alpha-1} \cdot d x \\ & =2^{4 \alpha}=2^8 \\ \alpha & =2 \end{aligned} $

Given, $U(x)=\left(5 x^2-4 x^3\right) \mathrm{J}$

We know, $F=\frac{-d U(x)}{d x}$

$ \begin{array}{ll} \Rightarrow & F=\frac{-d}{d x}\left(5 x^2-4 x^3\right) \\ & =-10 x+12 x^2 \\ \because & F=0 \,\,\,(given)\\ \therefore & 12 x^2-10 x=0 \\ \Rightarrow & 2 x(6 x-5)=0 \\ \therefore & 2 x=0 \text { or } 6 x-5=0 \\ \Rightarrow & x=0 \text { or } x=\frac{5}{6} \mathrm{~m} \end{array} $

According to law of conservation of energy at point $B$ shown in the figure given in question part, Loss in PE $=$ Gain in KE

$\Rightarrow \quad m g(H-h)=\frac{1}{2} m v^2$

$\Rightarrow \quad v=\sqrt{2 g(H-h)}$

Now, $\quad h=\frac{1}{2} g t^2 \Rightarrow t=\sqrt{\frac{2 h}{g}}$

$\therefore$ Distance covered in horizontal portion,

$\begin{aligned} s & =v \times t \\ & =\sqrt{2 g(H-h)} \times \sqrt{\frac{2 h}{g}} \\ \Rightarrow \quad s & =\sqrt{4 h(H-h)} \quad \text{... (i)} \end{aligned}$

For maximum value of $s$

$\begin{array}{l} \qquad\frac{d s}{d h}=0 \\ \Rightarrow \frac{1}{2 \sqrt{4 h(H-h)}} \times 4[(H-h) 1+h(-1)]=0 \\ \Rightarrow 2(H-2 h)=0 \Rightarrow H=2 h \\ \Rightarrow h=\frac{H}{2} \end{array}$

Substituting the value of $h$ in Eq. (i), we get,

$\begin{aligned} s & =\sqrt{4 \frac{H}{2}\left(H-\frac{H}{2}\right)} \\ & =\sqrt{2 H \cdot \frac{H}{2}}=H \end{aligned}$

Mass of block, $m=50 \mathrm{~kg}$,

speed of block, $v=4 \mathrm{~ms}^{-1}$,

$F=500 \mathrm{~N} \text { and } \theta=30^{\circ} \text {, }$

The rate at which the force does work on the block is equal to power which is given as

$\begin{aligned} P & =\text { Force in the horizontal direction } \times \text { velocity } \\ & =500 \times \cos 30^{\circ} \times V \\ & =500 \times \frac{\sqrt{3}}{2} \times 4=100 \sqrt{3} \\ & =100 \times 1.732=1732 \mathrm{~W} \end{aligned}$

Mass of ball, $m=300 \mathrm{~g}=0.3 \mathrm{~kg}$M

height, $h=10 \mathrm{~m}$

If $v$ be the speed of ball when it reaches the ground, then

$\begin{aligned} \quad v^2 & =u^2+2 g h=0+2 \times 10 \times 10 \\ \Rightarrow \quad v & =200 \Rightarrow v=10 \sqrt{2} \mathrm{~m} / \mathrm{s} \end{aligned}$

For sandy ground, initial velocity of ball,

$v=10 \sqrt{2} \mathrm{~m} / \mathrm{s}$

Final velocity $v^{\prime}=0$

distance travelled by the ball in sand, $s=1.5 \mathrm{~m}$ if $a^{\prime}$ be retardation applied on the ball in sandy ground, then using equation

$\begin{aligned} v^{\prime 2} & =v^2-2 a s \\ 0 & =(10 \sqrt{2})^2-2 a \times 1.5 \Rightarrow a=\frac{200}{3} \mathrm{~m} / \mathrm{s}^2 \end{aligned}$

$\because$ Average resistance offered by the sand

$F=m \cdot a=0.3 \times \frac{200}{3}=20 \mathrm{~N}$

Given, height, $h=1 \mathrm{~m}$

mass, $m=1 \mathrm{~kg}$

spring constant, $k=15 \mathrm{~Nm}^{-1}$

The kinetic energy of mass after falling on the platform will be converted into potential energy of the spring.

Thus, $m g(h+x)=\frac{1}{2} \mathrm{k} x^2$,

$\qquad 1 \times 10(1+x)=\frac{1}{2} \times 15 \times x^2$

$\begin{aligned} & \Rightarrow \quad 10 \times \frac{2}{15}(1+x)=x^2 \\ & \Rightarrow \quad \frac{4}{3}(1+x)=x^2 \\ & \Rightarrow \quad 4+4 x=3 x^2 \\ & \Rightarrow \quad 3 x^2-4 x-4=0 \\ & \Rightarrow \quad 3 x^2-6 x+2 x-4=0 \\ & \Rightarrow \quad 3 x(x-2)+2(x-2)=0 \\ & \Rightarrow \quad(x-2)=0 \\ & \text { or } \quad (3 x+2)=0 \\ & x=2 \mathrm{~m} \text { or } x=\frac{-2}{3} \mathrm{~m} \\ \end{aligned}$

Since, $x$ is the compression, which cannot be negative, thus $x=2 \mathrm{~m}$.

Given, mass of bead, $m=400 \mathrm{~g}=0.4 \mathrm{~kg}$

Power delivered, $P=1.2 \mathrm{~W}$

Time, $t=6 \mathrm{~s}$

when the force is applied, it gives energy to the body. So from work energy theorem,

$\begin{aligned} W & =\frac{1}{2} m v^2-\frac{1}{2} m u^2 \\ & =\frac{1}{2} m v^2 \end{aligned}$

Now, Power = Work/Time

$\begin{aligned} & \qquad P=W / t \\ & \Rightarrow \quad P \times t=W \\ & \Rightarrow \quad 1.2 \times 6=\frac{1}{2} \times m v^2 \\ & \Rightarrow \quad 7.2=\frac{0.4}{2} \times v^2 \\ \end{aligned}$

$\begin{array}{ll} \Rightarrow & v^2=36 \\ \Rightarrow & v=6 \mathrm{~ms}^{-1} \end{array}$

work done by the gun = total kinetic energy of the bullets

$ = n{1 \over 2}m{v^2} = 300 \times {1 \over 2} \times 10 \times {10^{ - 3}} \times {(600)^2}$

$ = 150 \times 10 \times {10^{ - 3}} \times 600 \times 600$

Power of gun $ = {{work\,done} \over {Time\,taken}}$

$ = {{150 \times 10 \times {{10}^{ - 3}} \times 600 \times 600} \over {60\,s}}$

= 9 kW ($\because$ 1 minute = 60 seconds)

Given, power developed by engine, $P=20 \mathrm{~kW}$

Mass, $m=200 \mathrm{~kg}$

Height, $H=40 \mathrm{~m}$

Acceleration due to gravity, $g=10 \mathrm{~ms}^{-1}$

$\text { Since, } \begin{aligned} P & =\frac{\text { energy }}{\text { time }}=\frac{m g h}{t} \\ t & =\frac{m g h}{P}=\frac{200 \times 10 \times 40}{20000}=4 \mathrm{~s} \end{aligned}$

Given, ratio of kinetic energies $E_1: E_2=4: 1$

Ratio of linear velocity, $v_1: v_2=1: 1$

Let ratio of masses be $m_1: m_2$

As we know that,

$\begin{aligned} E & =\frac{p^2}{2 m} \\ \therefore \quad \frac{E_1}{E_2} & =\left(\frac{p_1}{p_2}\right)^2\left(\frac{2 m_2}{2 m_1}\right)=\left(\frac{m_1 v_1}{m_2 v_2}\right)^2\left(\frac{m_2}{m_1}\right) \\ & =\frac{m_1^2}{m_2^2} \times \frac{m_2}{m_1}=\frac{m_1}{m_2}=\frac{4}{1} \\ \Rightarrow m_1: m_2 & =4: 1 \end{aligned}$

Given, height through which water falls $H=25 \mathrm{~m}$

Mass of water, $m=3 \times 10^3 \mathrm{~kg}$

Time, $t=60 \mathrm{~s}$

Let power be $P$.

As we know that,

$\begin{aligned} P & =\frac{\operatorname{Energy~}(E)}{\operatorname{Time}(t)} \\ \Rightarrow \quad P & =\frac{m g H}{t} \\ & =\frac{3 \times 10^3 \times 10 \times 25}{60}=12500 \mathrm{~W} \end{aligned}$

Given, range of projectile, $R=100 \mathrm{~m}$

Let initial and final velocities be $u$ and $v$ and angle of projection be $\theta$.

As we know that,

In case of projectile motion, the final velocity of body always less than that of initial velocity of projection $(v < u)$ and kinetic energy $(\mathrm{KE})=\frac{1}{2} m v^2$ where, $m$ is mass and $v$ is velocity.

$\therefore$ Kinetic energy will be again maximum, when it will be again on ground at $100 \mathrm{~m}$ from point of projection.

Given,

Force on body, $\mathbf{F}=17-2 x+6 x^2$

Mass of body, $m=2 \mathrm{~kg}$

Initial position, $x_i=0 \mathrm{~m}$

Final position, $x_f=8 \mathrm{~m}$

Since,

$\begin{aligned} & \text { work, } \int d W=\int \mathbf{F} \cdot \mathbf{d x} \\ & =\int\left(17-2 x+6 x^2\right) d x \\ & \Rightarrow \quad W=17 x-\frac{2 x^2}{2}+\left.\frac{6 x^3}{3}\right|_0 ^8 \\ & =17 x-x^2+\left.2 x^3\right|_0 ^8 \\ & =17 \times 8-8^2+2 \times 8^3 \\ & =1096 \mathrm{~J} \\ & \end{aligned}$

Let initial velocity $=u$

$\begin{aligned} \text { Final velocity after passing one plank } & =u-\frac{u}{25} \\ & =\frac{24}{25} u \end{aligned}$

Number of planks $=N$

Final velocity of bullet, $v=0 \mathrm{~ms}^{-1}$

Acceleration of bullet $=a$

As we know that,

$\begin{aligned} v^2-u^2 & =2 a s \\ \therefore \quad a & =\frac{v^2-u^2}{2 s} \\ & =\frac{\left(\frac{24}{25} u\right)^2-u^2}{2 s}=-\frac{49 u^2}{625 \times 2 s} \end{aligned}$

Again for '$N$' number of planks,

$\begin{aligned} & 0^2-u^2=2\left(\frac{49 u^2}{625 \times 2 s}\right) \cdot N s \\ \Rightarrow \quad N & =\frac{625}{49}=12.7 \approx 13 \end{aligned}$

Given,

Mass of chain = m

Length of chain = $l$

According to question,

Mass of chain hanging $=\frac{m}{6}$

Centre of mass of hanged part of chain,

$h=\frac{1}{2}\times\frac{l}{6}=\frac{l}{12}$

As we know that,

Work = $mgh$

where, g is acceleration due to gravity and h is height moved by chain.

$\therefore \quad W=\frac{m}{6}g\times\frac{l}{12}=\frac{mgl}{72}$

As we know that,

$\begin{aligned} & \text { Kinetic energy }(\mathrm{KE})=\frac{1}{2} \times \operatorname{mass}(M) \times[\text { velocity }(v)]^2 \\ & \Rightarrow \quad \mathrm{KE} \propto v^2 \end{aligned}$

As, by equation of parabola, $y=m x^2+c$

Equation is $\mathrm{KE}=\frac{1}{2} M v^2+0$

$\therefore$ Graph will be a parabola.

Given, power of motor,

$\begin{aligned} P & =\frac{1}{4} \mathrm{hp} \\ & =\frac{1}{4} \times 746 \mathrm{~W} \end{aligned}$

Angular speed of motor

$\begin{aligned} \omega & =600 \mathrm{rpm} \\ & =600 \times \frac{2 \pi}{60}=20 \pi ~\mathrm{rad~s}^{-1} \end{aligned}$

$\therefore$ Time for one rev $=\frac{60}{600}=0.1 \mathrm{~s}$

Efficiency, $\eta=\frac{60}{100}$

$\begin{aligned} & \because \quad P=\frac{W}{t} \\ & \therefore W(\text { in l rotation) }=P \times t \\ & \quad=\frac{746}{4} \times \frac{1}{10}=18.65 \mathrm{~J} \end{aligned}$

Hence, work done by motor

$\begin{aligned} & =\eta \times W=\frac{60}{100} \times 18.65 \\ & =11.19 \mathrm{~J} \end{aligned}$

Given, mass of body, $m=8 \mathrm{~kg}$

Equation of displacement, $s=t^2 / 4$

Time, $t=4 \mathrm{~s}$

As we know that,

$W =\mathbf{F} \cdot \mathbf{s}$

$=m \mathbf{a} \cdot \mathbf{s}$ ...... (i)

where, $W$ is work, F is force, $\mathbf{a}$ is acceleration and $\mathbf{s}$ is displacement.

Now, $a=\frac{d^2 s}{d t^2}=\frac{d}{d t}\left(\frac{2 t}{4}\right)=\frac{2}{4}=\frac{1}{2} \mathrm{~ms}^{-2}$

and $s=\frac{4^2}{4}=\frac{16}{4}=4 \mathrm{~m} $

Now, put the value of $m, a$ and $S$ in Eq. (i), we get

$W=8 \times \frac{1}{2} \times 4=16 \mathrm{~J} $

Work done by the particle in X-direction is given by

dW = F dx = (a + bx) dx

Total work done in displacement from x = 0 to x = d will be

$W = \int {dW = \int\limits_0^d {(a + bx)dx} } $

$ = \left[ {ax + {{b{x^2}} \over 2}} \right]_0^d$

$ \Rightarrow W = \left( {a + {{bd} \over 2}} \right)d$

Kinetic energy at Q = 90% of potential energy at P

$ \Rightarrow {1 \over 2}m{v^2} = {{90} \over {100}} \times mgh$

$ \Rightarrow v = \sqrt {1.8\,gh} \Rightarrow v = \sqrt {1.8 \times 10 \times 2} $

$ \Rightarrow v = 6$ m/s

Here, applied force, $F=10 \mathrm{~N}$

Friction force, $f=\mu N=0.2 \times m g$

$ =0.2 \times 2 \times 10=4 \mathrm{~N} $

Net force, $F_{\text {net }}=F-f=10-4=6 \mathrm{~N}$

∴ Acceleration, $a=\frac{F_{\text {net }}}{m}=\frac{6}{2}=3 \mathrm{~ms}^{-2}$

Distance covered in 4s,

$ s=0+\frac{1}{2} a t^2=\frac{1}{2} \times 3 \times(4)^2=24 \mathrm{~m} $

Work done by applied force of 10 N ,

$ W_1=F s=10 \times 24=240 \mathrm{~J} $

Work done by friction force of 4 N ,

$ \begin{aligned} W_2 & =f s \cos 180^{\circ}=-4 \times 24 \\ & =-96 \mathrm{~J} \end{aligned} $

Given, $x=\frac{\alpha t^2}{2}$

As we know that, $v=\frac{d x}{d t}$

$ \begin{equation*} =\frac{d}{d t}\left(\frac{\alpha t^2}{2}\right)=\alpha t \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{equation*} $

According to work-energy theorem

Work done, $W=\frac{1}{2} m v^2$

$ \begin{aligned} =\frac{1}{2} m( & \alpha t)^2 \quad \quad \quad \text { [using Eq. (i)] } \\ & =\frac{1}{2} m \alpha^2 t^2 \\ & =\frac{1}{2} \times 2 \times(1)^2 \times(2)^2=4 \mathrm{~J} \end{aligned} $

The law of conservation of energy is valid both, macroscopic and microscopic domain.

All conserved quantities are not necessarily scalars. e.g., conservation of linear momentum and angular momentum are vectors whereas conservation of energy is scalar.

Given, mass of particle,

$ m=30 \mathrm{~g}=3 \times 10^{-2} \mathrm{~kg} $

Displacement of the particle is given as

$ x=\alpha t^2 $

$ \begin{array}{ll} \therefore \text { Velocity, } & v=\frac{d x}{d t}=\frac{d}{d t} \alpha t^2 \\ \Rightarrow & v=2 \alpha t \end{array} $

Acceleration, $a=\frac{d v}{d t}=\frac{d}{d t}(2 \alpha t) \Rightarrow a=2 \alpha$

∴ Force on the particle,

$ \begin{aligned} F & =m a=3 \times 10^{-2} \times 2 \alpha \\ & =6 \times 10^{-2} \mathrm{~N} \quad[\because \alpha=1] \end{aligned} $

$ \begin{aligned} &\text { ∴ Work done during first } 4 \mathrm{~s} \text {, }\\ &\begin{aligned} W & =\int_0^4 F d x=\int_0^4 6 \times 10^{-2} d x \\ & =6 \times 10^{-2} \int_0^4 d x \\ & =6 \times 10^{-2} \int_0^4 v d t \quad[\because d x=v d t] \\ & =6 \times 10^{-2} \int_0^4 2 \alpha t d t \\ & =6 \times 10^{-2} \times 2 \alpha \cdot\left[\frac{t^2}{2}\right]_0^4 \\ & =12 \times 10^{-2} \times 8 \quad[\because \alpha=1] \\ & =0.96 \mathrm{~J} \end{aligned} \end{aligned} $

Displacement of object is along $X$-axis by 3 m ,

$ \mathbf{x}=3 \hat{\mathbf{i}} \mathrm{~m} $

$ \text { Force }=2 \hat{\mathbf{i}}+4 \hat{\mathbf{j}} \mathrm{~N} $

Work done by force, $W=\mathbf{F} \cdot \mathbf{x}$

$ =(2 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}) \cdot 3 \hat{\mathbf{i}}=6 \mathrm{~J} $

$ \mathbf{u}=20 \hat{\mathbf{i}}+24 \hat{\mathbf{j}} $

To find change in PE, we have to consider only vertical motion of the ball.

So, we have

$ \begin{aligned} & u_y=24 \mathrm{~m} / \mathrm{s}, m=1 \mathrm{~kg} \\ & a_y=-10 \mathrm{~m} / \mathrm{s}^2, t=6 \mathrm{~s} \end{aligned} $

We take origin at point of projection,

$ \begin{aligned} h & =u_y t-\frac{1}{2} g t^2 \\ & =24 \times 6-\frac{1}{2} \times 10 \times 36 \\ & =144-180=-36 \mathrm{~m} \end{aligned} $

So, vertical displacement of the particle is

$ h=-36 \mathrm{~m} $

Hence, change in PE $=m g h$

$ =1 \times 10 \times-36=-360 \mathrm{~J} $

By work energy theorem, Change in $\mathrm{KE}=$ Work done by net (resultant) force on the body

Given, displacement, $x=\frac{t^3}{25}$

So, velocity, $v=\frac{d x}{d t}=\frac{3 t^2}{25} \mathrm{~ms}^{-1}$

Acceleration, $a=\frac{d v}{d t}=\frac{6 t}{25} \mathrm{~ms}^{-2}$

Force $=$ Mass × Acceleration

$ F=10 \cdot\left(\frac{6 t}{25}\right)=\frac{12 t}{5} \mathrm{~N} $

Also,   $ \frac{d x}{d t}=\frac{3 t^2}{25} \Rightarrow d x=\frac{3 t^2}{25} \cdot d t $

Work done in displacing object by displacement $d x$ is

$ d W=F \cdot d x=\frac{12 t}{5} \cdot \frac{3 t^2}{25} \cdot d t=\frac{36}{5 \times 25} t^3 d t \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

So, total work done by force in first 5 s is

$ \begin{aligned} W & =\int_0^5 d W=\int_0^5 \frac{36}{5 \times 25} \times t^3 d t \text { [from Eq. (i)] } \\ & =\frac{36}{5 \times 25} \times\left[\frac{t^4}{4}\right]_0^5 \\ & =\frac{36 \times(5)^4}{5 \times 25 \times 4}=9 \times 5=45 \mathrm{~J} \end{aligned} $

Given, initial speed of ball, v = 16 ms^$-$1

Maximum height, h = 9 m

Mass of ball, m = 0.5 kg

Energy supplied to ball while throwing up,

${E_1} = {1 \over 2}m{v^2} = {1 \over 2} \times 0.5 \times 16 \times 16 = 64$ J

Energy stored when ball reaches the maximum height,

E₂ = mgh = 0.5 $\times$ 9.8 $\times$ 9 = 44.1 J

Energy dissipated by air drag acting on the ball during the ascent,

E = E₁ = E₂ = 64 $-$ 44.1 = 19.9 J

Given, force, F = (8t) N, Mass, m = 2 kg

From Newton's second law,

Force = Rate of change of momentum

$F = {{dp} \over {dt}}$

i.e. dp = Fdt

$\therefore$, $p = \int {dp = \int_0^1 {F\,.\,dt = \int_0^1 {8t\,.\,dt} } } $

$ = 8\left[ {{{{t^2}} \over 2}} \right]_0^1 = 8\left( {{1 \over 2} - 0} \right)$

$\Rightarrow$ p = 4 kg ms^$-$1

Change in kinetic energy, $\Delta K = {{{p^2}} \over {2m}} = {{{{(4)}^2}} \over {2 \times 2}} = 4$ J

From work-energy theorem,

Work done = Change in kinetic energy

So, work done by the force during the first 1s,

W = $\Delta$K = 4J

Work done, W = area under F-s graph

$\therefore$ Total work done is covering an initial distance of 40 m,

W = area of trapezium ABCD + area of trapezium CEFD

$ = {1 \over 2} \times 20(20 + 30) + {1 \over 2} \times 10 \times (20 + 40) = 800$ J

[$\because$ Area of trapezium = ${1 \over 2}$ $\times$ height $\times$ sum of parallel sides]

$\Rightarrow$ W = 800 J