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 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} $.

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 $.