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} $

$ \text { } \begin{aligned} W & =\int_0^5 F d x \\ & =\int_0^5\left(3 x^2-2 x+7\right) d x \\ & =\left[3 \cdot \frac{x^3}{3}-\frac{2 x^2}{2}+7 x\right]_0^5 \\ & =\left[x^3-x^2+7 x\right]_0^5 \\ & =\left[5^3-5^2+7 \times 5-0+0-0\right] \\ & =135 \mathrm{~J} \end{aligned} $

$ \begin{aligned} W & =\mathbf{F} \cdot \mathbf{s} \\ & =(3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}) \cdot(2 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+1 \hat{\mathbf{k}}) \\ & =3 \times 2+2 \times 2+5 \times 1 \\ & =15 \mathbf{J} \end{aligned} $

Potential energy of a body of at a height of $5 \mathrm{~m}=100 \mathrm{~J}$

$ \begin{aligned} m g h & =100 \\ \Rightarrow \quad m & =\frac{100}{g h}=\frac{100}{10 \times 5}=2 \mathrm{~kg} \end{aligned} $

At height $h=10 \mathrm{~m}$, velocity of the body is given as

$ \begin{aligned} v^2 & =u^2-2 g h \\ & =20^2-2 \times 10 \times 10=200 \\ \therefore \quad \mathrm{KE} & =\frac{1}{2} m v^2=\frac{1}{2} \times 2 \times 200 \\ & =200 \mathrm{~J} \end{aligned} $

$ \text { } \begin{aligned} \text { } & W=F \cdot s \\ = & (2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}) \\ & \cdot[(2-3) \hat{\mathbf{i}}+(2-0) \hat{\mathbf{j}}+(3+4) \hat{\mathbf{k}}] \\ = & (2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}) \cdot[-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+7 \hat{\mathbf{k}}] \\ = & -2+6+28=32 \mathrm{~J} \end{aligned} $

$ \begin{aligned} \text { Power, } & P=\frac{W}{t}=\frac{m g h}{t} \\ = & \frac{7560 \times 10 \times 100}{60 \times 60}=2100 \mathrm{~W} \end{aligned} $

Since, $\eta=0.7$

$ \begin{aligned} \Rightarrow \frac{P_{\text {out }}}{P_{\text {in }}} & =0.7 \\ \Rightarrow \quad P_i & =\frac{P_{\text {out }}}{0.7}=\frac{2100}{0.7} \\ & =3000 \mathrm{~W}=3 \mathrm{~kW} \end{aligned} $

$ \begin{aligned} & U=5 x(x-4)=5 x^2-20 x \\ & \frac{d U}{d x}=10 x-20 \end{aligned} $

$U$ may be maximum or minimum if

$ \begin{aligned} & \frac{d U}{d x}=0 \\ & 10 x-20=0 \\ & x=2 \mathrm{~m} \\ & \frac{d^2 u}{d x^2}=10 \end{aligned} $

At $x=2 \mathrm{~m}, \frac{d^2 U}{d x^2}=10(+\mathrm{ve})$

Thus, $U$ is minimum at $x=2 \mathrm{~m}$.

According to conservation of energy, kinetic energy will be maximum at $x=2 \mathrm{~m}$.

Thus, speed of particle is maximum at $x=2 \mathrm{~m}$

Displacement vector of body is

$ \begin{aligned} \mathbf{S} & =\left(x_2-x_1\right) \hat{\mathbf{i}}+\left(y_2-y_1\right) \hat{\mathbf{j}} \\ & =(2-1) \hat{\mathbf{i}}+(0-2) \hat{\mathbf{j}} \\ & =1 \hat{\mathbf{i}}-2 \hat{\mathbf{j}} \mathrm{~m} \end{aligned} $

Force on body

$ \mathbf{F}=(3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}) \mathrm{N} $

Work done, $W=\mathbf{F} \cdot \mathbf{S}$

$ \begin{aligned} & =(3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}) \cdot((\hat{\mathbf{i}}-2 \hat{\mathbf{j}}) \\ & =3+4=7 \mathrm{~N} \cdot \mathrm{~m} \\ & =7 \mathrm{~J} \end{aligned} $

$ \begin{aligned} &\text { }\\ &\begin{aligned} W & =\int_{x_1}^{x_2} f d x \\ & =\int_2^4\left(6 x^2-4 x\right) d x \\ & =\left[\frac{6 x^3}{3}-\frac{4 x^2}{2}\right]_2^4 \\ & =\left[2 x^3-2 x^2\right]_2^4 \\ & =2 \times 4^3-2 \times 4^2-2 \times 2^3+2 \times 2^2 \\ & =128-32-16+8 \\ & =88 \mathrm{~J} \end{aligned} \end{aligned} $

Step 1: Find the volume of water in the well

The well is a cylinder. Its diameter is 2 m, so its radius $ r = 1 $ m. Its depth (height) $ h = 14 $ m. The volume formula for a cylinder is $ V = \pi r^2 h $. So, the volume of water is $ V = \pi \times 1^2 \times 14 = 14\pi $ cubic meters.

Step 2: Find the mass of the water

The density ($ \rho $) of water is $ 1000 $ kg/m³. So, mass ($ m $) = volume × density = $ 14\pi \times 1000 $ kg.

Step 3: Calculate work needed to lift all the water

The work to lift water from the bottom to the ground level is $ W = mgh $, where $ g = 10 $ m/s² and $ h $ is the height (14 m). So, $ W = (1000 \times 14 \pi) \times 10 \times 7 $

Step 4: Substitute values and calculate

$ W = 1000 \times \frac{22}{7} \times 1^2 \times 14 \times 10 \times 7 = 3.08 \times 10^6 \text{ J} $

Step 5: Use the power of the motor to find time

Power ($ P $) is $ 1.4 $ kW, which is $ 1.4 \times 10^3 $ W.

Time ($ t $) = $ \frac{W}{P} = \frac{3.08 \times 10^6}{1.4 \times 10^3} $.

$ t = 22 \times 10^2 \text{ s } = 2200 \text{ s} $

$ \begin{aligned} &\text { According to work energy theorem }\\ &\begin{aligned} W & =\frac{1}{2} m v^2 \\ \therefore \quad P & =\frac{W}{t} \\ P & =\frac{1}{2} m v^2 / t \Rightarrow v=\sqrt{\frac{2 P t}{m}} \\ & =\sqrt{\frac{2 \times 10 \times 10^3 \times 10}{2000}}=10 \mathrm{~m} / \mathrm{s} \end{aligned} \end{aligned} $

The body moves at a steady speed, which means its speed does not change. This only happens if the total (net) force on the body is zero.

$ \begin{array}{ll} \text{So,} & F_1 - F_2 = 0 \\ \text{which means} & F_1 = F_2 \end{array} $

Work and Energy:

Because the body's speed is not changing, its kinetic energy also stays the same. The work-energy theorem tells us that the total work done on the body is equal to the change in its kinetic energy.

$ \begin{aligned} & W = \Delta E_k \\ & W = 0 \end{aligned} $

Net Power:

Power is the work done divided by time. Since the total work ($W$) is zero, the net power is also zero:

$ P = \frac{W}{t} = 0 $

Let's use the work-energy theorem, which says:

$ \begin{aligned} & & K & =W \\ \Rightarrow & & K & =F \times s \quad ...(i) \end{aligned} $

Here, $K$ is kinetic energy, $F$ is force, and $s$ is distance.

When a force acts and the body starts from rest, the distance $s$ covered in time $t$ is:

$ s = \frac{1}{2} a t^2 $

Since acceleration $a = \frac{F}{m}$ (where $m$ is mass), we get:

$ s = \frac{1}{2} \times \frac{F}{m} \times t^2 \quad ...(ii) $

Now, substitute $s$ from (ii) into our equation (i):

$ K = F \times \frac{1}{2} \times \frac{F}{m} \times t^2 $

So, $K = \frac{F^2}{2m} t^2$

We are told both bodies $A$ and $B$ get the same kinetic energy ($K_A = K_B$) using the same force $F=40$ N, but their masses and times are different.

Set up the equation for both bodies and compare:

$ \frac{F^2}{2m_A} t_A^2 = \frac{F^2}{2m_B} t_B^2 $

You can cancel out $F^2$ and $\frac{1}{2}$ because they are the same:

$ \frac{t_A^2}{m_A} = \frac{t_B^2}{m_B} $

Rearrange to solve for time ratio:

$ \frac{t_A^2}{t_B^2} = \frac{m_A}{m_B} $

Take the square root for the ratio $t_A : t_B$:

$ \frac{t_A}{t_B} = \sqrt{\frac{m_A}{m_B}} $

Plug in $m_A = 20$ kg and $m_B = 5$ kg:

$ \frac{t_A}{t_B} = \sqrt{\frac{20}{5}} = \sqrt{4} = 2 $

So, the ratio $t_A : t_B = 2 : 1$.

Potential energy gained by crane,

$ \begin{aligned} U & =m g h \\ & =8000 \times 10 \times 108 \\ & =8.64 \times 10^6 \mathrm{~J} \end{aligned} $

$\therefore$ Output power of crane

$ \begin{aligned} P_0 & =\frac{W}{t}=\frac{U}{t}=\frac{8.64 \times 10^6}{60 \times 60} \\ \Rightarrow \quad P_0 & =2.4 \times 10^3 \mathrm{~W}=2.4 \mathrm{~kW} \end{aligned} $

Since, efficiency of crane

$ \begin{aligned} \eta & =\frac{P_0}{P_i} \\ \Rightarrow \quad 0.8 & =\frac{P_0}{P_i} \\ \Rightarrow \quad P_i & =\frac{P_0}{0.8} \\ & =\frac{2.4}{0.8} \mathrm{~kW}=3 \mathrm{~kW} \end{aligned} $

$v=108 \mathrm{~km} / \mathrm{h}=108 \times \frac{5}{18}=30 \mathrm{~m} / \mathrm{s}$

Total frictional force on the train

$ f_s=\frac{10^6 \times 0.5}{100}=5000 \mathrm{~N} $

$\therefore$ Power of the train

$ \begin{aligned} P & =f_s \cdot v=5000 \times 30 \\ & =150000 \mathrm{~W}=150 \mathrm{~kW} \end{aligned} $

Step 1: Write the equation for power

Power ($P$) is force ($F$) multiplied by velocity ($v$):

$P = F \cdot v$

Step 2: Replace force with mass and acceleration

Force is mass ($m$) times acceleration ($\frac{dv}{dt}$):

$P = m \frac{dv}{dt} \cdot v$

Step 3: Rearrange to separate variables

Move terms so you have all $v$ terms on one side and $t$ terms on the other:

$v \frac{dv}{dt} = \frac{P}{m}$

Step 4: Integrate both sides

Integrate with respect to $v$ and $t$:

$\int v \, dv = \frac{P}{m} \int dt$

After integrating:

$\frac{v^2}{2} = \frac{P}{m} t$

Step 5: Solve for velocity ($v$)

$v = \sqrt{\frac{2P t}{m}}$

Step 6: Find the relationship between displacement and time

Velocity is the rate of change of displacement, so:

$\frac{ds}{dt} = \sqrt{\frac{2P t}{m}}$

Step 7: Integrate to get displacement ($s$)

Integrate both sides with respect to $t$:

$\int ds = \sqrt{\frac{2P}{m}} \int t^{1/2} dt$

The right side integrates to $t^{3/2}$:

$s \propto t^{3/2}$

Step 8: Answer

The value of $x$ is $\frac{3}{2}$.

Initial kinetic energy

$ \begin{aligned} E_{k_1} & =\frac{1}{2} m v^2 \\ \Rightarrow \quad X & =\frac{1}{2} m v^2 \end{aligned} $

Kinetic energy at highest point

$ \begin{aligned} E_{k_2} & =\frac{1}{2} m\left(v \cos 60^{\circ}\right)^2 \\ & =\frac{1}{2} m v^2\left(\frac{1}{2}\right)^2=\frac{1}{4}\left(\frac{1}{2} m v^2\right) \\ & =\frac{1}{4}(X) \\ & =\frac{X}{4} \end{aligned} $

Given Data

• Length of pendulum, $ l = 200 \text{ cm} = 2 \text{ m} $

• Released from horizontal position → Initial height (relative to mean position) = $ l = 2 \text{ m} $

• Acceleration due to gravity, $ g = 10 \text{ m/s}^2 $

• $10\%$ of energy lost due to air resistance

Step 1: Initial Potential Energy

At horizontal position (bob just released):

$ E_{\text{initial}} = m g h = m g l = m(10)(2) = 20m \text{ J} $

Step 2: Energy Lost in Air Resistance

$10\% $ of energy is lost → only $90\%$ remains.

$ E_{\text{mean}} = 0.9 \times E_{\text{initial}} = 0.9 \times 20m = 18m \text{ J} $

Step 3: Energy at Mean Position

At mean position, the bob has only kinetic energy (potential energy = 0):

$ \frac{1}{2} m v^2 = 18m $

$ v^2 = 36 \quad \Rightarrow \quad v = 6 \text{ m/s} $

✅ Final Answer:

$ \boxed{v = 6~\text{m/s}} $

Correct Option: A) $6~\mathrm{m/s}$

Frictional force

$ \begin{aligned} f_s & =\mu m g \\ & =0.05 \times 3 \times 10^6 \times 10 \\ & =1.5 \times 10^6 \mathrm{~N} \end{aligned} $

$\therefore$ Power, $P=f_s v$

$ \begin{aligned} & =1.5 \times 10^6 \times 50 \\ & =75 \times 10^6 \mathrm{~W} \\ & =75 \mathrm{WM} \end{aligned} $

$p=m v=24 \mathrm{~kg} \mathrm{~m} / \mathrm{s}$

$ \therefore u=\frac{24}{m}=\frac{24}{8}=3 \mathrm{~m} / \mathrm{s} $

Impulse, $J=F \cdot \Delta t$

$ \begin{aligned} & =24 \times 3=72 \mathrm{Ns} \\ \therefore \quad J & =\Delta p=72 \mathrm{Ns} \end{aligned} $

$\therefore$ Final momentum

$ \begin{aligned} p_f & =p_i+\Delta p=24+72 \\ & =96 \mathrm{~kg} \mathrm{~m} / \mathrm{s} \\ \therefore \quad v_f & =\frac{p_f}{m}=\frac{96}{8}=12 \mathrm{~m} / \mathrm{s} \end{aligned} $

Thus, increase in kinetic energy,

$ \begin{aligned} \Delta K & =K_f-K_i \\ & =\frac{1}{2} m v_f^2-\frac{1}{2} m u^2 \\ & =\frac{1}{2} m\left(v_f^2-u^2\right) \\ & =\frac{1}{2} \times 8\left(12^2-3^2\right) \\ & =4(135)=540 \mathrm{~J} \end{aligned} $

According to work energy theorem,

Work done, $W=\Delta K$

$ \begin{aligned} & =K_f-K_i \\ & =\frac{1}{2} \times 0.25 \times 12^2-\frac{1}{2} \times 0.25 \times 0 \\ & =18 \mathrm{~J} \end{aligned} $

If $F$ be the net force acted on the ball, then

$ \begin{aligned} & W=F \cdot s \\ \Rightarrow \quad & F=\frac{W}{s}=\frac{18}{0.9}=20 \mathrm{~N} \end{aligned} $

Work done in stretching a string to obtain an extension l,

$ W_1=\frac{1}{2} k l^2 $

Similarly, work done in stretching a string to obtain an extension $l_1$ is,

$ W_2=\frac{1}{2} k l_1^2 $

∴ Work done in second case, $W$

$ =W_2-W_1=\frac{1}{2} k\left(l_1^2-l^2\right) $

$ \begin{aligned} & \text {As, } F=\mu m g \cos \theta \\ & \text { or } \quad F=0.30 \times 10 \times 10 \cos 45^{\circ} \\ & \text { or } \quad F=\frac{30}{\sqrt{2}} \mathrm{~N} \end{aligned} $

$ \begin{aligned} W & =F \times s \times \cos 180^{\circ}=-\frac{30}{\sqrt{2}} \times 5 \\ & =-\frac{150}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}=-75 \sqrt{2} \mathrm{~J} \end{aligned} $

As the vertical component of the velocity at highest point is zero.

$ u_y=0, u_x=u \cos \beta $

The kinetic energy at the highest point will be equal to $\frac{1}{2} m(u \cos \beta)^2$

The initial kinetic energy $=\frac{1}{2} m u^2$

$\therefore \quad \frac{1}{2} m(u \cos \beta)^2=\frac{3}{4} \times \frac{1}{2} m u^2$

$\Rightarrow \quad \cos \beta=\frac{\sqrt{3}}{2} \Rightarrow \beta=30^{\circ}$

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} $

To determine the average power imparted by friction before the particle stops, consider the following:

Mass and Initial Conditions:

The particle has a mass $ m $ and is initially at rest before being given a velocity $ u $.

The surface has a coefficient of friction $ \mu $.

Work Done by Friction:

The work done by friction $ W_f $ is accounted by the change in kinetic energy as the particle stops. This can be equated to the initial kinetic energy:

$ W_f = \frac{1}{2} m u^2 $

Acceleration Due to Friction:

The deceleration $ a $ due to friction is given by the frictional force divided by the mass:

$ a = \frac{\mu m g}{m} = \mu g $

Time to Stop:

The time $ t $ taken for the particle to come to a rest can be calculated from the equation of motion $ v = u + at $, setting final velocity $ v = 0 $:

$ 0 = u - at \quad \Rightarrow \quad t = \frac{u}{a} = \frac{u}{\mu g} $

Average Power:

The average power $ P $ imparted by the friction is the total work done over the time taken to stop:

$ P = \frac{W_f}{t} = \frac{\frac{1}{2} m u^2}{\frac{u}{\mu g}} = \frac{1}{2} m u \mu g $

Thus, the average power imparted by friction before the particle stops is $\frac{1}{2} \mu m g u$.

Given:

Force: $\mathbf{F} = (4 \hat{\mathbf{i}} + 2 \hat{\mathbf{j}} + \hat{\mathbf{k}}) \, \mathrm{N}$

Initial position: $\mathbf{r}_1 = (2 \hat{\mathbf{i}} + 2 \hat{\mathbf{j}} + \hat{\mathbf{k}}) \, \mathrm{m}$

Final position: $\mathbf{r}_2 = (4 \hat{\mathbf{i}} + 3 \hat{\mathbf{j}} + 2 \hat{\mathbf{k}}) \, \mathrm{m}$

Calculating Total Displacement:

The displacement $\mathbf{s}$ is given by the difference between the final and initial positions:

$ \begin{aligned} \mathbf{s} & = \mathbf{r}_2 - \mathbf{r}_1 \\ & = (4 \hat{\mathbf{i}} + 3 \hat{\mathbf{j}} + 2 \hat{\mathbf{k}}) - (2 \hat{\mathbf{i}} + 2 \hat{\mathbf{j}} + \hat{\mathbf{k}}) \\ & = (2 \hat{\mathbf{i}} + \hat{\mathbf{j}} + \hat{\mathbf{k}}) \end{aligned} $

Calculating Work Done:

The work done by the force is calculated using the dot product of the force vector and the displacement vector:

$ \begin{aligned} W & = \mathbf{F} \cdot \mathbf{s} \\ & = (4 \hat{\mathbf{i}} + 2 \hat{\mathbf{j}} + \hat{\mathbf{k}}) \cdot (2 \hat{\mathbf{i}} + \hat{\mathbf{j}} + \hat{\mathbf{k}}) \\ & = (4 \times 2) + (2 \times 1) + (1 \times 1) \\ & = 8 + 2 + 1 \\ & = 11 \, \mathrm{J} \end{aligned} $

Thus, the work done by the force on the particle is 11 Joules.

When a body is thrown vertically upwards and reaches a maximum height $ H $, we can analyze the situation using conservation of energy. The ratio of velocities at specified heights involves comparing kinetic and potential energies at those points:

Maximum Height $ H $:

Potential Energy (PE) at the maximum height is given by:

$ \text{PE}_{\max} = mgH $

Height $ \frac{3H}{4} $:

Potential Energy at this height:

$ \text{PE}_1 = mg\left(\frac{3H}{4}\right) $

Kinetic Energy (KE) at this height using energy conservation:

$ \text{KE}_1 = \text{PE}_{\max} - \text{PE}_1 = mgH - mg\left(\frac{3H}{4}\right) = mg\left(\frac{H}{4}\right) $

Height $ \frac{8H}{9} $:

Potential Energy at this height:

$ \text{PE}_2 = mg\left(\frac{8H}{9}\right) $

Kinetic Energy at this height:

$ \text{KE}_2 = \text{PE}_{\max} - \text{PE}_2 = mgH - mg\left(\frac{8H}{9}\right) = mg\left(\frac{H}{9}\right) $

Velocities $ v_1 $ and $ v_2 $:

Velocity at $ \frac{3H}{4} $:

$ v_1 = \sqrt{\frac{2\text{KE}_1}{m}} = \sqrt{2g\left(\frac{H}{4}\right)} $

Velocity at $ \frac{8H}{9} $:

$ v_2 = \sqrt{\frac{2\text{KE}_2}{m}} = \sqrt{2g\left(\frac{H}{9}\right)} $

Ratio of Velocities:

The ratio of the velocities is:

$ \frac{v_1}{v_2} = \frac{\sqrt{2gH/4}}{\sqrt{2gH/9}} = \sqrt{\frac{9}{4}} = \frac{3}{2} $

Thus, the ratio of the velocities of the body at the heights $\frac{3H}{4}$ and $\frac{8H}{9}$ is $3:2$.

Given:

Mass (m): 90 kg

Acceleration due to gravity (g): 10 m/s²

Height climbed (h): 600 m

Time (t): 90 minutes = 90 × 60 = 5400 seconds

To find the average power generated by the climber, we first calculate the work done using the formula for work:

$ W = m \times g \times h $

Substituting the given values:

$ W = 90 \times 10 \times 600 = 540,000 \, \text{Joules (J)} $

Next, the average power ($ P_{\text{av}} $) can be calculated using the formula:

$ P_{\text{av}} = \frac{W}{t} $

Substitute the known values:

$ P_{\text{av}} = \frac{540,000}{5400} $

$ P_{\text{av}} = 100 \, \text{Watts (W)} $

Thus, the average power generated by the climber is 100 Watts.

Given:

Light body momentum: $ p_L $

Heavy body momentum: $ p_H $

The kinetic energy for an object is given by:

$ K = \frac{p^2}{2m} $

For the lighter body:

The kinetic energy $ K_L $ is:

$ K_L = \frac{p_L^2}{2m_L} $

For the heavier body:

The kinetic energy $ K_H $ is:

$ K_H = \frac{p_H^2}{2m_H} $

Since both bodies have the same kinetic energy:

$ \frac{p_L^2}{2m_L} = \frac{p_H^2}{2m_H} $

From this equation, we can derive:

$ p_H^2 = \frac{m_H}{m_L} p_L^2 $

Given that the mass of the heavy body ($ m_H $) is greater than that of the light body ($ m_L $), the ratio $ \frac{m_H}{m_L} $ is greater than one. Therefore, it follows that:

$ p_H > p_L $

To determine the velocity of a block when it reaches the ground, given a machine with an efficiency of $\frac{2}{3}$:

Calculate the Work Done:

Energy Used: 12 J

Efficiency: $\frac{2}{3}$

Work done $= \text{Efficiency} \times \text{Energy}$

$ \text{Work done} = \frac{2}{3} \times 12 = 8 \text{ J} $

Apply the Work-Energy Theorem:

Work Done: Equals the change in potential energy.

$ W = mgh $

Given:

Mass $m = 2 \, \text{kg}$

Gravitational acceleration $g = 10 \, \text{m/s}^2$

$ 8 = 2 \times 10 \times h $

$ h = \frac{8}{20} \, \text{m} = 0.4 \, \text{m} $

Use Conservation of Energy:

The potential energy when the block is lifted is converted to kinetic energy when it falls.

$ \frac{1}{2} mv^2 = mgh $

Solve for $v:$

$ v = \sqrt{2gh} $

Substitute the values:

$ v = \sqrt{2 \times 10 \times \frac{8}{20}} $

$ v = \sqrt{2 \times 10 \times 0.4} $

$ v = \sqrt{8} $

$ v = 2\sqrt{2} \, \text{m/s} $

Thus, the velocity of the block when it reaches the ground is $2 \sqrt{2} \, \text{m/s}$.

Let's consider an inclined plane with a total length of $ L $, where the top $\frac{1}{n}$ of the plane is smooth, and the remaining lower part is rough with a coefficient of friction $\mu_k$. A body starts from rest at the top and returns to rest at the bottom.

Length of the smooth part: $\frac{L}{n}$

Length of the rough part: $\frac{n-1}{n} \cdot L$

Angle of inclination: $\theta$

Coefficient of friction: $\mu_k$

Motion on the Smooth Part

The acceleration down the inclined plane due to gravity is $ a = g \sin \theta $.

The distance traveled on this smooth part is:

$ s_1 = \frac{L}{n} $

Using the kinematic equation for motion:

$ v_1^2 = u^2 + 2as_1 $

$ v_1^2 = 0 + 2(g \sin \theta) \frac{L}{n} = \frac{2gL \sin \theta}{n} $

Motion on the Rough Part

The acceleration on this part is:

$ a' = g \sin \theta - \mu_k g \cos \theta $

The distance traveled here is:

$ s_2 = \frac{n-1}{n} \cdot L $

Final velocity at the end of this segment ($ v_2 $) is 0, and the initial velocity ($ u_2 $) is $ v_1 $.

Using the equation for this motion:

$ v_2^2 = u_2^2 + 2a's_2 $

$ 0 = v_1^2 + 2a's_2 $

Substitute the expression for $ a' $:

$ 0 = \frac{2gL \sin \theta}{n} + 2(g \sin \theta - \mu_k g \cos \theta) \cdot \frac{n-1}{n} \cdot L $

Simplify to solve for $\theta$:

$ 0 = 2gL \sin \theta + 2gL(n-1)(\sin \theta - \mu_k \cos \theta) $

$ 0 = \sin \theta + (n-1)(\sin \theta - \mu_k \cos \theta) $

$ 0 = n \sin \theta - (n-1)\mu_k \cos \theta $

$ \Rightarrow n \sin \theta = (n-1)\mu_k \cos \theta $

$ \Rightarrow \tan \theta = \frac{(n-1)\mu_k}{n} $

Thus, the angle of inclination $\theta$ is:

$ \theta = \tan^{-1}\left[\frac{(n-1) \mu_k}{n}\right] $

Given the spring constant $ k = 200 \, \mathrm{N/m} $.

Initial stretch $ x_1 = 10 \, \mathrm{cm} = 0.1 \, \mathrm{m} $

Total (final) stretch $ x_2 = 20 \, \mathrm{cm} = 0.2 \, \mathrm{m} $

To find the work done to further stretch the spring by 10 cm, we use the formula for work done on a spring:

$ W = U_2 - U_1 $

Where:

$ U_2 $ is the elastic potential energy at the final stretch

$ U_1 $ is the elastic potential energy at the initial stretch

The formula for elastic potential energy $ U $ stored in a spring is:

$ U = \frac{1}{2} k x^2 $

Substituting for $ W $:

$ W = \frac{1}{2} k x_2^2 - \frac{1}{2} k x_1^2 $

Plug in the values:

$ W = \frac{1}{2} \times 200 \, \mathrm{N/m} \times \left[(0.2 \, \mathrm{m})^2 - (0.1 \, \mathrm{m})^2\right] $

$ W = 100 \times (0.04 - 0.01) $

$ W = 100 \times 0.03 $

$ W = 3 \, \mathrm{J} $

Therefore, the work done to stretch the spring an additional 10 cm is $ 3 \, \mathrm{J} $.

FBD of pulley $A$

So, $\quad T=m g \Rightarrow T_1=T+m g$

$ T_1=T+m g=2 m g $

$ \text { FBD of pulley } B $

We get, $\quad 2 T_1=k x$

$ 2(2 m g)=k x \Rightarrow x=\frac{4 m g}{k} $

where $x$ is the extension in the spring,

$ \begin{aligned} x & =\frac{4 \times 4 \times 10}{8 \times 10^3} \\ & =2 \times 10^{-2} \mathrm{~m}=2 \mathrm{~cm} \end{aligned} $

Given two bodies, $ A $ and $ B $, with masses and initial velocities as follows:

Mass of body $ A $, $ M_A = 2m $

Mass of body $ B $, $ M_B = m $

Initial velocity of $ A $, $ u_A = u $

Initial velocity of $ B $, $ u_B = 2u $

The maximum height $ H_A $ reached by body $ A $ can be calculated using the formula for maximum height in vertical motion:

$ H_A = \frac{u_A^2}{2g} = \frac{u^2}{2g} $

The height $ h $, which is half of the maximum height $ H_A $ reached by body $ A $, is given by:

$ h = \frac{H_A}{2} = \frac{\frac{u^2}{2g}}{2} = \frac{u^2}{4g} $

Using the conservation of energy principle, the velocity $ v_A $ of body $ A $ at height $ h $ is determined as follows:

$ \begin{aligned} & \frac{1}{2} M_A u_A^2 = \frac{1}{2} M_A v_A^2 + M_A g h \\ & \Rightarrow \frac{1}{2}(2m) u^2 = \frac{1}{2}(2m) v_A^2 + (2m) g \left(\frac{u^2}{4g}\right) \\ & \Rightarrow u^2 = v_A^2 + \frac{u^2}{2} \\ & \Rightarrow v_A = \frac{u}{\sqrt{2}} \end{aligned} $

The kinetic energy $ K_A $ of body $ A $ at height $ h $ is:

$ \begin{aligned} K_A & = \frac{1}{2} M_A v_A^2 \\ & = \frac{1}{2}(2m)\left(\frac{u}{\sqrt{2}}\right)^2 \\ & = \frac{1}{2}(2m)\left(\frac{u^2}{2}\right) \\ & = \frac{mu^2}{2} \end{aligned} $

The potential energy $ P_B $ of body $ B $ at height $ h $ is:

$ P_B = M_B g h = m g \left(\frac{u^2}{4g}\right) = \frac{mu^2}{4} $

Finally, the ratio of the kinetic energy of body $ A $ to the potential energy of body $ B $ is calculated as follows:

$ \begin{aligned} \text{Ratio} & = \frac{K_A}{P_B} \\ & = \frac{\frac{mu^2}{2}}{\frac{mu^2}{4}} \\ & = \frac{2}{1} \\ K_A : P_B & = 2:1 \end{aligned} $

To calculate the work done by the force, we use the formula for work done by a variable force:

$ W = \int_{x_1}^{x_2} F \, dx $

Given that the force $ F(x) = 6x^2 - 4x + 3 $ and the displacement is from $ x = 2 \, \text{m} $ to $ x = 5 \, \text{m} $, the work done is:

$ \begin{aligned} W &= \int_2^5 (6x^2 - 4x + 3) \, dx \\ &= \left[ \frac{6x^3}{3} - \frac{4x^2}{2} + 3x \right]_2^5 \\ &= \left[ 2x^3 - 2x^2 + 3x \right]_2^5 \\ &= \left[ 2 \times 5^3 - 2 \times 5^2 + 3 \times 5 \right] - \left[ 2 \times 2^3 - 2 \times 2^2 + 3 \times 2 \right] \\ &= (250 - 50 + 15) - (16 - 8 + 6) \\ &= 215 - 14 \\ &= 201 \, \text{J} \end{aligned} $

Thus, the work done by the force is $ 201 \, \text{J} $.

Given:

Spring constant, $ k = 5 \times 10^3 \, \mathrm{N/m} $

Initially stretched position, $ x_1 = 10 \, \mathrm{cm} = 0.1 \, \mathrm{m} $

Further stretched position, $ x_2 = 20 \, \mathrm{cm} = 0.2 \, \mathrm{m} $

Work Done:

The work done to further stretch the spring from $ x_1 $ to $ x_2 $ is equal to the change in potential energy of the spring. This can be calculated using the formula:

$ \text{Work done} = \frac{1}{2} k \left[\left(x_2\right)^2 - \left(x_1\right)^2\right] $

Substitute the given values:

$ \begin{aligned} & = \frac{1}{2} \times 5 \times 10^3 \left[(0.2)^2 - (0.1)^2\right] \\ & = \frac{1}{2} \times 5 \times 10^3 \times (0.04 - 0.01) \\ & = \frac{1}{2} \times 5 \times 10^3 \times 0.03 \\ & = 2.5 \times 10^3 \times 0.03 \\ & = 75 \, \mathrm{N \cdot m} \end{aligned} $

Therefore, the work required to stretch the spring further by another 10 cm is $ 75 \, \mathrm{N \cdot m} $.

To determine the work done by the lower block on the upper block, follow these steps:

Calculate Maximum Possible Acceleration:

The maximum acceleration the upper block (1 kg) can experience without sliding off is due to static friction:

$ a_{\text{max}} = \mu g = 1 \, \text{m/s}^2 $

where $\mu$ is the coefficient of friction and $g$ is the acceleration due to gravity.

Calculate Common Acceleration:

The applied force is distributed over the total mass (3 kg) of the system, so the acceleration $a$ is:

$ a = \frac{F}{m_{\text{total}}} = \frac{0.5}{3} = \frac{0.5}{3} \, \text{m/s}^2 $

Check for Relative Motion:

Since $ a < a_{\text{max}} $, the blocks will move together without slipping.

Calculate the Force of Friction:

The force of friction acting from the lower block to the upper block is:

$ f = m_{\text{upper}} \cdot a = 1 \cdot \frac{0.5}{3} = \frac{1}{6} \, \text{N} $

Calculate the Work Done by Friction:

The work done by this friction force over a displacement of 3 meters is:

$ W = f \times s = \frac{1}{6} \times 3 = 0.5 \, \text{J} $

Thus, the work done by the lower block on the upper block for a displacement of 3 meters is $0.5 \, \text{J}$.

Power of tension $ =0 $

Power of $ m g=(m g)(v) \cos 60^{\circ} $

Here, $ \quad v=\sqrt{2 g h} $

and $ \quad h=l\left(\cos 30^{\circ}-\cos 60^{\circ}\right)=0.36 \mathrm{~m} $

$ v=\sqrt{2 \times 10 \times 0.36} $

$ =2.68 \mathrm{~m} / \mathrm{s} $

Power $ =(m g)(v) \cos 60^{\circ} $

$ \begin{array}{l} =(1 \times 10)(268) \times \frac{1}{2} \\ =1.34 \times 10=13.4 \mathrm{~W} \end{array} $