Work, Energy and Power

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.

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

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

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

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} $

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} $