Work, Energy and Power

$ \begin{aligned} & \text { Power }=\frac{\text { Work }}{\text { Time }} \\ & =\frac{m g h}{t} \\ & =\frac{10^3 \times 9.8 \times 20}{10}=19.6 \mathrm{~kW} \end{aligned} $

Given $\mu_k(x)=\mu_0-\alpha x$ and $\mu_k(L)=0=\mu_0-\alpha L \Rightarrow \mu_0=\alpha L$ Magnitude of work done by frictional force, $W_f=\int d W_f$ where $d W_f=f_k d x=\mu_k N d x=\mu_k M g d x=\left(\mu_0-\alpha x\right) M g d x$

$ \begin{aligned} & \therefore W_f=\int_0^L \mu_0 M g d x-\int_0^L \alpha M g x d x=\mu_0 M g[x]_0^L-\alpha M g\left[\frac{x^2}{2}\right]_0^L \\ & \Rightarrow n \mu_0 M g L=\mu_0 M g L-\left(\frac{\mu_0}{L}\right) M g\left(\frac{L^2}{2}\right)=\frac{\mu_0 M g L}{2} \\ & \Rightarrow n=\frac{1}{2} \end{aligned} $

According to the work-energy theorem, the work done by the braking force is equal to the change in kinetic energy. Thus, for each car, we have:

$ FS = \Delta KE $

This can be rearranged to:

$ -FS = k_f - k_i \quad \Rightarrow \quad FS = k_i - k_f $

For both cars, since the final kinetic energy $ k_f $ is zero when they stop, the equation simplifies to:

$ FS = k_i $

For car $ A $ and car $ B $, the forces $ F_A $ and $ F_B $ applied by the brakes satisfy:

$ F_A \cdot S_A = k_A \quad \text{and} \quad F_B \cdot S_B = k_B $

Solving for the ratio of the forces:

$ \frac{F_A}{F_B} = \frac{k_A}{k_B} \times \frac{S_B}{S_A} $

Substituting the given kinetic energies and stopping distances:

$ \frac{F_A}{F_B} = \frac{100}{225} \times \frac{1500}{1000} $

This simplifies to:

$ = \frac{150}{225} = \frac{2}{3} $

Thus, the ratio of the forces $ \frac{F_A}{F_B} $ is $ \frac{2}{3} $.

At Point $P, m g \sin \theta=\frac{m v^2}{l}\quad\text{..... (1)}$

By conservation of mechanical energy at point $P \in Q$

$\frac{1}{2} m v_0^2=\frac{1}{2} m v^2+m g(I+I \sin \theta)$

$\begin{aligned} & \frac{v_0^2}{2}=\frac{v^2}{2}+g l(1+\sin \theta) \\ & \text { Put gl }=\frac{v^2}{\sin \theta} \text { using (1) } \\ & \frac{v_0^2}{2}=\frac{v^2}{2}+\frac{v^2}{\sin \theta}(1+\sin \theta) \\ & \frac{v_0^2}{2}=\frac{v^2}{2}+\frac{v^2}{\sin \theta}+v^2 \\ & \frac{v_0^2}{2}=\frac{3}{2} v^2+\frac{2 v^2}{2 \sin \theta} \\ & v_0^2=v^2\left[3+\frac{2}{\sin \theta}\right] \\ & \frac{v}{v_0}=\left(\frac{\sin \theta}{3 \sin \theta+2}\right)^{\frac{1}{2}} \end{aligned}$

To find the kinetic energy of the object at the end of the displacement, we need to calculate the work done by the force $\vec{F}$ on the object during its displacement. The work done by a force is given by the dot product of the force and the displacement vectors:

$ W = \vec{F} \cdot \vec{d} $

Here, the force $\vec{F}$ is given as:

$ \vec{F} = -2 \hat{i} + 3 \hat{j} \, \mathrm{N} $

And the displacement $\vec{d}$ is given as:

$ \vec{d} = 3 \hat{i} \, \mathrm{m} $

The dot product of the two vectors is calculated as follows:

$ W = (-2 \hat{i} + 3 \hat{j}) \cdot (3 \hat{i}) $

Since the $\hat{j}$ component of the force does not contribute to the work done in the $x$-direction, it can be ignored. Thus, the dot product yields:

$ W = -2 \cdot 3 \, \mathrm{N} \cdot \mathrm{m} = -6 \, \mathrm{J} $

The negative sign indicates that the force does work against the direction of displacement. Now, the initial kinetic energy of the object is:

$ K_i = 10 \, \mathrm{J} $

The work-energy theorem relates the work done on an object to its change in kinetic energy:

$ W = K_f - K_i $

Rearranging for the final kinetic energy $K_f$ gives:

$ K_f = K_i + W $

Substituting the known values:

$ K_f = 10 \, \mathrm{J} + (-6 \, \mathrm{J}) $

This simplifies to:

$ K_f = 4 \, \mathrm{J} $

Hence, the kinetic energy of the object at the end of the displacement is Option C: $4 \mathrm{~J}$.

To solve this problem, we need to consider the principles of energy conservation and the behavior of the object during the rebound. Let's break it down step-by-step:

1. When the object falls from a height of $10 \mathrm{~m}$, it converts its potential energy to kinetic energy at the point of impact. The potential energy (PE) just before hitting the ground can be calculated using the formula:

$PE = mgh$

where:

• $m$ is the mass of the object
• $g$ is the acceleration due to gravity ($9.8 \mathrm{~m/s^2}$)
• $h$ is the height ($10 \mathrm{~m}$)

The kinetic energy (KE) of the object just before impact is equal to the potential energy it had at the height of $10 \mathrm{~m}$, because of the conservation of energy principle.

2. Upon hitting the ground, the object loses $50\%$ of its kinetic energy. Therefore, the kinetic energy after the impact is:

$KE_{after\ impact} = \frac{1}{2} KE_{before\ impact}$

3. The object will now rebound to a height where its kinetic energy is converted back to potential energy. Let this height be $h_{rebound}$. The potential energy at this rebound height is:

$PE_{rebound} = mgh_{rebound}$

Since the kinetic energy after the impact is $\frac{1}{2} mgh$, we set this equal to the potential energy at the rebound height:

$\frac{1}{2} mgh = mgh_{rebound}$

By canceling out the mass $m$ and the gravitational constant $g$, we get:

$\frac{1}{2} h = h_{rebound}$

Substituting $h = 10 \mathrm{~m}$, we find:

$\frac{1}{2} \times 10 \mathrm{~m} = h_{rebound}$

$h_{rebound} = 5 \mathrm{~m}$

Therefore, the height up to which the object can rebound from the ground is $5 \mathrm{~m}$. The correct answer is:

Option D $5 \mathrm{~m}$

To find the instantaneous power delivered by a force, we can use the formula:

$ P = \vec{F} \cdot \vec{v} $

where $ P $ is the power, $ \vec{F} $ is the force vector, and $ \vec{v} $ is the velocity vector of the particle.

In this question, the force exerted is given as a constant $ 5 \, N $. Since no direction is specified, and the displacement is given in a scalar form, we assume the force acts along the direction of displacement. Therefore, we can treat the vectors as scalars for simplicity.

The displacement of the particle is given by:

$ x = 2t - 1 $

The velocity, $ \vec{v} $, is the derivative of displacement with respect to time. Deriving the displacement equation with respect to time $ t $ gives:

$ v = \frac{dx}{dt} = \frac{d}{dt}(2t - 1) = 2 $

The instantaneous power can now be calculated as:

$ P = F \cdot v = 5 \times 2 = 10 \, \text{Watts} $

Thus, the instantaneous power delivered by the force at any time $ t $ is $ 10 \, \text{Watts} $.

The correct answer is Option A: 10.

$\begin{aligned} & P=\vec{F} \cdot \vec{V} \\ & P=(10 \hat{i}+10 \hat{j}+20 \hat{k}) \cdot(5 \hat{i}-3 \hat{j}+6 \hat{k}) \\ & P=50-30+120 \\ & P=140 \mathrm{~W} \end{aligned}$

The potential energy (U) stored in a spring when it is stretched or compressed is given by the formula:

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

where:

• $k$ is the spring constant,
• $x$ is the displacement from the spring's equilibrium position (i.e., how much the spring is stretched or compressed).

If the spring is initially stretched by $2$ cm (or $0.02$ meters, since we generally use SI units for these calculations), the potential energy stored can be represented as:

$U = \frac{1}{2} k (0.02)^2$

When the spring is stretched by $8$ cm (or $0.08$ meters), the new potential energy $U'$ becomes:

$U' = \frac{1}{2} k (0.08)^2$

To find the relation between $U'$ and $U$, we can divide the expression for $U'$ by that of $U$:

$\frac{U'}{U} = \frac{\frac{1}{2} k (0.08)^2}{\frac{1}{2} k (0.02)^2}$

Simplifying this quotient gives:

$\frac{U'}{U} = \left(\frac{0.08}{0.02}\right)^2 = \left(4\right)^2 = 16$

Thus, if the spring's displacement is increased from $2$ cm to $8$ cm, the potential energy stored in the spring increases by a factor of $16$, meaning:

$U' = 16U$

$F \propto - x$

$F = - kx$

Where F is restoring force

x is displacement of block from equilibrium position

F_up = 2000 g + 3000

= 23000 N

Minimum power P_min = $\overrightarrow F $ . $\overrightarrow v $

P_min = Fv = 23000 $\times$ ${3 \over 2}$

= 34500 W

Energy = Power $\times$ time

E = 100 $\times$ 10³ $\times$ 3600

= 36 $\times$ 10⁷ J

Take the surface of the Earth as the reference level for potential energy.

Initially, the particle is released from height $S$, so

• Initial potential energy $= mgS$

• Initial kinetic energy $= 0$

So, total mechanical energy is

$ E = mgS $

Now suppose at some instant the particle is at height $h$ above the surface.

Then,

• Potential energy at that instant $= mgh$

• Kinetic energy at that instant is given to be three times the potential energy:

$ K = 3U = 3mgh $

Using conservation of mechanical energy:

$ K + U = mgS $

Substitute $K = 3mgh$ and $U = mgh$:

$ 3mgh + mgh = mgS $

$ 4mgh = mgS $

$ h = \frac{S}{4} $

So the height from the surface is

$ \frac{S}{4} $

Now find the speed at that instant.

Since

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

and also

$ K = 3mgh $

So,

$ \frac{1}{2}mv^2 = 3mgh $

Cancel $m$:

$ \frac{1}{2}v^2 = 3gh $

$ v^2 = 6gh $

Now put $h = \frac{S}{4}$:

$ v^2 = 6g \cdot \frac{S}{4} = \frac{3gS}{2} $

Therefore,

$ v = \sqrt{\frac{3gS}{2}} $

Hence, the correct answer is:

$ \boxed{\frac{S}{4},\ \sqrt{\frac{3gS}{2}}} $

So, Option A is correct.

Given :

Mass of the block, $ m $ = 0.25 kg

From the work-energy theorem, the potential energy of the spring equals the potential energy of the block :

$ \frac{1}{2} k x^2 = m g x $

Where :

$ k $ is the force constant of the spring

$ x $ is the compression in the spring

Solving for $ kx $ we get :

$ kx = 2mg $

Substituting $ m = 0.25 \, \text{kg} $ :

$ kx = 2 \times 0.25 \times g $

$ kx = 0.5g $

The force applied by the spring on the block is :

$ F = kx \Rightarrow F = 0.5g $

From the free body diagram :

Thus, the maximum force $ N $ exerted by the system on the floor is :

$ N = F + 2g \\ N = 0.5g + 2g \\ N = 2.5g \\ N = 25 \, \text{N} $

Free body diagram of block is shown below.

Now, from the energy conservation,

$\begin{aligned} w & =\Delta K \\\\ \Rightarrow \quad w_F+w_{s p} & =\frac{1}{2} m v^2-0 \\\\ \therefore \quad F_x-\frac{1}{2} k x^2 & =\frac{1}{2} m v^2 \\\\ v & =\sqrt{\frac{2 F_x-k x^2}{m}} \end{aligned}$

For the first displacement $y=0$

Hence, $F_x=0$ and no work is done. For the second displacement, $F_y=-k a$ and $\Delta y=a$

Work $=F_y \Delta y=-k a^2$

Let at any instant of time, the length $A B$ be $l$. Here angle $\theta$ and length $l$ vary with time, then using Pythagorus theorem in $\triangle A B C$

Differentiating both sides w.r.t. '$t$', we get

$x^2+y^2=l^2$

Differentiating both sides w.r.t. '$t$', we get

or $\quad 2 x \frac{d x}{d t}+2 y \frac{d y}{d t}=2 l \frac{d l}{d t}$

As, there no vertical motion of the block, so

$\begin{aligned} & & \frac{d y}{d t} & =0, \frac{d x}{d t}=v_x \text { and } \frac{d l}{d t}=v \\ & \therefore & 2 x v_x & =2 l v \\ & \text { or } & v_x & =\frac{l}{x} v \\ & \text { or } & v_x & =\frac{v}{\left(\frac{x}{l}\right)}=\frac{v}{\sin \theta} \end{aligned}$

Given, $m=70 \mathrm{~kg}, g=10 \mathrm{~ms}^{-2}, h=12 \mathrm{~m}$ In going up and down once.

Number of k-cals burnt $=\left(m g h+\frac{m g h}{2}\right)$

$\begin{aligned} & =\frac{3}{2} m g h \\ & =\frac{3}{2} \times \frac{70 \times 10 \times 12}{4.2 \times 1000} \\ & =3 \mathrm{k}-\mathrm{cal} \end{aligned}$

Total no. of k-cal to be burnt for loosing $7 \mathrm{~kg}$ of weight $=7 \times 9000=63000 \mathrm{~k}$-cal

$\therefore$ Number of times the person has to go up and down the stairs

$\begin{aligned} & =\frac{63000}{3}=21000 \\ & =21 \times 10^3 \text { times } \end{aligned}$