Work, Energy and Power

$ \begin{aligned} &\\ &\begin{aligned} & \text {} \mathbf{V}=(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-4 \hat{\mathbf{k}})(\mathrm{m} / \mathrm{s}) \\ & \therefore v=\sqrt{2^2+3^2+(-4)^2}=\sqrt{29} \mathrm{~m} / \mathrm{s} \\ & K=87 \mathrm{~J} \\ & \Rightarrow \frac{1}{2} m v^2=87 \\ & \Rightarrow \frac{1}{2} \times m \times 29=87 \Rightarrow m=6 \mathrm{~kg} \end{aligned} \end{aligned} $

We know that Power, $P=F \cdot v=\operatorname{mav}=m \cdot \frac{d v}{d t} \cdot v$

$ P=m v \frac{d v}{d t} $

$ \begin{aligned} & \Rightarrow\left(3 t^2+3\right)=m v \frac{d v}{d t} \\ & \Rightarrow\left(3 t^2+3\right) d t=0.5 v d v \\ & \Rightarrow v d v=\left(6 t^2+6\right) d t \\ & \Rightarrow \int_0^v v d v=\int_0^3\left(6 t^2+6\right) d t \\ & \Rightarrow\left[\frac{v^2}{2}\right]_0^v=\left[6 \cdot \frac{t^3}{3}+6 t\right]_0^3 \\ & \Rightarrow \frac{v^2}{2}=\left[2 t^3+6 t\right]_0^3=2 \times 3^3+6 \times 3-0 \\ & \quad=72 \\ & \therefore v^2=72 \times 2=144 \Rightarrow v=12 \mathrm{~m} / \mathrm{s} \end{aligned} $

$ \begin{aligned} &\text { Work done }\\ &\begin{aligned} W & =\int F d y=\int_a^{2 a} \frac{-K}{y^2} d y=\left[\frac{K}{y}\right]_a^{2 a} \\ & =\left[\frac{K}{2 a}-\frac{K}{a}\right]=\frac{-K}{2 a} \end{aligned} \end{aligned} $

Initial energy of body

$ E_i=m g h=0.5 \times 10 \times 3.2=16 \mathrm{~J} $

Final energy of the body

$ E_f=\frac{1}{2} m v^2=\frac{1}{2} \times 0.5 \times 6^2=9 \mathrm{~J} $

∴ Loss of energy due to air resistance

$ \Delta U=U_i-U_f=16-9=7 \mathrm{~J} $

Work done by the force in displacing the body is given by,

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

Given, Force vector, $ \mathbf{F}=(8 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}) $

Initial position vector, $ r_{1}=(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-4 \hat{\mathbf{k}}) $

Final position vector, $ \mathbf{r}_{2}=(4 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}) $

The displacement vector $ s $ is given by

$ \begin{array}{l} \mathbf{s}=\mathbf{r}_{2}-\mathbf{r}_{1} \\\\ \mathbf{s}=2 \hat{\mathbf{i}}-6 \hat{\mathbf{j}}+10 \hat{\mathbf{k}} \end{array} $

Putting the value of Fand $ s $ in Eq. (i), we get

$ \begin{array}{l} W=(8 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}) \cdot(2 \hat{\mathbf{i}}-6 \hat{\mathbf{j}}+10 \hat{\mathbf{k}}) \\\\ W=16+12+60 \\\\ W=88 \mathrm{~J} \end{array} $

Given:

Mass $ m = 4 \, \text{kg} $

Velocity $ \mathbf{v} = (2 \hat{\mathbf{i}} - 4 \hat{\mathbf{j}} - \hat{\mathbf{k}}) \, \text{m/s} $

The kinetic energy, $ \text{KE} $, of the body is calculated using the formula:

$ \text{KE} = \frac{1}{2} m v^2 $

To find $ v^2 $:

$ v^2 = (2 \hat{\mathbf{i}} - 4 \hat{\mathbf{j}} - \hat{\mathbf{k}}) \cdot (2 \hat{\mathbf{i}} - 4 \hat{\mathbf{j}} - \hat{\mathbf{k}}) $

$ v = \sqrt{2^2 + (-4)^2 + (-1)^2} $

$ v = \sqrt{4 + 16 + 1} = \sqrt{21} \, \text{m/s} $

Substituting the values back into the kinetic energy formula:

$ \text{KE} = \frac{1}{2} \times 4 \times (\sqrt{21})^2 $

$ = 2 \times 21 $

$ \text{KE} = 42 \, \text{J} $

Spring constant, $ k=100 \mathrm{~N} / \mathrm{m} $

Spring stretched, $ A=8 \mathrm{~cm} $

Position from mean where energy has $ x=3 \mathrm{~cm} $ to be find.

Kinetic energy is given as

$ \begin{aligned} \mathrm{KE} & =\frac{1}{2} k\left(A^{2}-x^{2}\right) \\\\ & =\frac{1}{2} \times 100\left(8^{2}-3\right)^{2} \times 10^{-4} \\\\ & =50(64-9) \times 10^{-4}=50 \times 55 \times 10^{-4} \\\\ \mathrm{KE} & =0.275 \mathrm{~J} \end{aligned} $

The work done in blowing a soap bubble is given by the formula $ W = T \cdot \Delta A $, where $ T $ represents the surface tension and $ \Delta A $ indicates the change in surface area.

The volume of a sphere (representing the soap bubble) is given by:

$ V = \frac{4}{3} \pi r^{3} $

where $ r $ is the radius of the soap bubble.

The surface area of the sphere is:

$ A = 4 \pi r^{2} $

From the above formulas, we can deduce that:

$ A \propto V^{\frac{2}{3}} $

Therefore, the change in area, $\Delta A$, is proportional to $(\Delta V)^{\frac{2}{3}}$.

To find the ratio of work done in expanding the bubble, we use:

$ \frac{W_{1}}{W_{2}} = \frac{\Delta A_{1}}{\Delta A_{2}} = \left(\frac{\Delta V_{1}}{\Delta V_{2}}\right)^{\frac{2}{3}} $

For a bubble expanding from volume $ V $ to volume $ 2V $:

$ W_{2} = W_{1}\left(\frac{2V}{V}\right)^{\frac{2}{3}} $

Substituting $ W_{1} = W $ (the original work done for volume $ V $):

$ W_{2} = W[2]^{\frac{2}{3}} $

Which simplifies to:

$ W_{2} = W(4)^{\frac{1}{3}} $

Body mass $ =m $

At maximum height $ h $

According to conservation of energy,

Total Energy (TE) $ = $ Potential Energy (PE)

Kinetic Energy (KE)

at maximum height velocity $ =.0 $

So,

$ \mathrm{TE}=\mathrm{PE}=m g h $

Now, at $ 40 \% $ of maximum height $ =\frac{2}{5} h $

$ (\mathrm{PE})_{40 \%}=m g\left(\frac{2}{5} h\right) $

Applying conservation of energy equation to find KE at $ 40 \% $ maximum height

$ \begin{aligned} (\mathrm{KE})_{40 \%} & =(\mathrm{TE})-(\mathrm{PE})_{40 \%} \\\\ & =m g h-m g\left(\frac{2}{5} h\right) \\\\ & =\frac{3 m g h}{5} \end{aligned} $

Taking the ratio of $ (\mathrm{KE})_{40 \%} $ and (PE) $ 40 \% $

$ \frac{(\mathrm{KE})_{40 \%}}{(\mathrm{PE})_{40 \%}}=\frac{\frac{3 m g h}{5}}{\frac{2 m g h}{5}}=\frac{3}{2}=3: 2 $

Let the body slide a distance $s$ along the incline before coming to rest.

So, it ascends by a vertical height $h=s \sin \theta$

From work-energy theorem,

$ \begin{array}{ll} & \Delta W_{f f}+\Delta W_{m g}=\Delta \mathrm{KE} \\\\ \Rightarrow \quad & -\mu m g(\cos \theta) s-m g s \sin \theta=0-E \\\\ \Rightarrow \quad & \mu m g s \cos \theta+m g s \sin \theta=E \end{array} $

$ \Rightarrow \quad s=\frac{E}{m g(\mu \cos \theta+\sin \theta)} $

So work done by friction,

$ \Delta W_{f f}=-\mu m g s \cos \theta=\frac{-\mu E \cos \theta}{\mu \cos \theta+\sin \theta} $

-ve sign is for direction.

The stopping distance of a moving object depends on its mass, velocity, and the force applied to stop it. The formula for stopping distance can be obtained from the work-energy principle:

$ F \cdot s = \frac{1}{2} m v^2 $

where:

$ F $ is the force applied by the brakes,

$ s $ is the stopping distance,

$ m $ is the mass of the moving object,

$ v $ is the velocity of the object.

Assuming the brakes apply a constant stopping force in both scenarios, we compare the initial and new stopping scenarios.

Initial Scenario:

Total mass $ m_1 = 80 \, \text{kg (mass of man)} + 100 \, \text{kg (mass of scooter)} = 180 \, \text{kg} $

Stopping distance: $ s_1 $

New Scenario (with bag of rice):

New total mass $ m_2 = 80 \, \text{kg (mass of man)} + 100 \, \text{kg (mass of scooter)} + 60 \, \text{kg (mass of rice)} = 240 \, \text{kg} $

New stopping distance: $ s_2 $

Analysis:

Using the work-energy principle for both scenarios and assuming velocity $ v $ and force $ F $ remain constant:

Initial condition:

$ F \cdot s_1 = \frac{1}{2} \times 180 \times v^2 $

New condition:

$ F \cdot s_2 = \frac{1}{2} \times 240 \times v^2 $

Dividing the second equation by the first:

$ \frac{s_2}{s_1} = \frac{\frac{1}{2} \times 240 \times v^2}{\frac{1}{2} \times 180 \times v^2} = \frac{240}{180} = \frac{4}{3} $

Rearranging gives:

$ s_2 = \frac{4}{3} s_1 $

or equivalently:

$ 3 s_2 = 4 s_1 $

Thus, the correct relationship is given by:

Option D: $ 4 s_1 = 3 s_2 $

Given, mass of block $=0.5 \mathrm{~kg}$

Kinetic friction, $\mu_k=0.2$

Horizontal force, $F=5 \mathrm{~N}$

$ t=4 \mathrm{~s} $

Frictional force,

$ \begin{aligned} & f_{\text {friction }}=\mu_k \times m \times g \\\\ & f_{\text {friction }}=1 \mathrm{~N} \end{aligned} $

Net force

$ \begin{aligned} & F_{\text {net }}=F_{\text {applied }}-f_{\text {friction }} \\\\ & F_{\text {net }}=4 \mathrm{~N} \end{aligned} $

Acceleration,

$ a=\frac{F_{\mathrm{nct}}}{m} $

$ \begin{array}{ll} \Rightarrow & a=8 \mathrm{~m} / \mathrm{s}^2 \\\\ \text { Velocity, } & v=u+a \times t \\\\ \Rightarrow & v=0+8 \times 4=32 \mathrm{~m} / \mathrm{s} \end{array} $

Kinetic energy,

$ \begin{aligned} \mathrm{KE} & =\frac{1}{2} m v^2 \\\\ & =\frac{1}{2} \times 0.5 \times 1024 \\\\ & =256 \mathrm{~J} \end{aligned} $

Kinetic energy at 4 s is 256 J .

Given, initial diameter $=2 \mathrm{~cm}$

Initial radius, $r_1=0.01 \mathrm{~m}$

Final diameter $=4 \mathrm{~cm}$

Final radius, $r_2=0.02 \mathrm{~m}$

Surface Tension, $T=3.5 \times 10^{-2} \mathrm{Nm}^{-1}$

Work done, $W=T \times \Delta A \quad \ldots \text { (i) }$

where, $\Delta A$ is change in surface area

$ \begin{aligned} \Delta A & =2 \times 4 \pi\left(r_2^2-r_1^2\right) \quad \ldots \text { (ii) } \\\\ & =2 \times 4 \pi\left[(0.02)^2-(0.01)^2\right] \\\\ \Delta A & =8 \pi(0.0003) \\\\ \Delta A & =2.4 \pi \times 10^{-3} \mathrm{~m}^2 \quad \ldots \text { (iii) } \end{aligned} $

Putting value of Eq. (iii) in Eq. (i)

Thus,

$ \begin{aligned} & W=3.5 \times 10^{-2} \times 2.4 \times \pi \times 10^{-3} \mathrm{~J} \\\\ & W=8.4 \pi \times 10^{-5} \\\\ & W=8.4 \times 3.14 \times 10^{-5}=26.3 \times 10^{-5} \\\\ & W=263 \times 10^{-6} \mathrm{~J} \end{aligned} $

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.

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} $

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} $