Q.Why is electrical power required at all when the elevator is descending? Why should there be a limit on the number of passengers in this case?

Convert the wind speed: $v = 36\ \text{kmph} = 36 \times \tfrac{5}{18} = 10\ \text{ms}^{-1}$.

The kinetic power carried by the wind passing through the swept area $A$ is

Substituting $\rho = 1.2\ \text{kg m}^{-3}$, $A = 20\ \text{m}^2$, $v = 10\ \text{ms}^{-1}$: …

The key idea is to separate thermal power (the raw energy released per second by fission) from output power (the useful electrical power after efficiency losses). The efficiency tells us what fraction of the thermal power is converted to electrical output.

1. Find the thermal power produced by fission. Each fission releases $200\ \text{MeV}$ of energy. The number of fissions per second is $5 \times 10^{13}$. So the energy released per second (thermal power) is:

Convert MeV to joules: $1\ \text{MeV} = 1.6 \times 10^{-13}\ \text{J}$.

Simplify:

So the reactor’s thermal power is $1600\ \text{W}$.

2. Apply the efficiency to get output power. Efficiency $\eta = 25\% = 0.25$. Output power is the useful electrical power: $$$$

Why this approach works

The key is to ask: What force actually does the work that increases the person’s gravitational potential energy?

Gravitational potential energy changes when the center of mass moves upward against gravity. The work done against gravity is $mgh$. But the force that directly lifts the person is not gravity — it’s the force the person’s legs exert on their own body. The stairs only provide a support (normal force), but that normal force does no net work on the person because the point of application (the foot) doesn’t move relative to the step while the force is applied (the step is stationary). So the real source is internal: the person’s muscles convert chemical energy into mechanical work.

Step-by-step reasoning

1. Identify the energy change

   As the person climbs, their gravitational potential energy increases by $\Delta U = mgh$. This energy must come from work done on the person by some force.

2. Check the normal force from the steps

   The normal force acts perpendicular to the step’s surface. While the foot is in contact, the step does not move — the normal force is applied at a stationary point. Work is $W = \vec{F} \cdot \Delta \vec{r}$, and here $\Delta \vec{r} = 0$ for the point of application. So the normal force does zero work.

   Watch out

   A common mistake is to think the normal force “pushes you up.” But the step doesn’t move with you — your foot pushes down on the step, and the step pushes back, but that push does no work because the contact point is stationary. The work is done by your muscles lifting your body.

3. Check the frictional force

   Friction prevents your foot from slipping, but it acts horizontally (parallel to the step surface). It does no vertical work. Even if friction helps you move forward, it doesn’t directly increase gravitational potential energy.

4. Check air resistance …

Why friction adjusts to what the car needs

Static friction is a responsive force. It doesn't always act at its maximum value $\mu_s N$; instead, it supplies exactly the force required to prevent slipping, up to that maximum limit. When a car rounds a curve, the tyres need an inward (centripetal) force to keep the car on the circular path. Static friction between the tyres and the road provides this force—as much as needed, but no more—until the demand exceeds $\mu_s N$, at which point the tyres slip.

The maximum speed tells us the radius of the track. At that speed, friction is pushed to its limit: $f_{\text{max}} = \mu_s m g = \frac{m v_{\text{max}}^2}{r}$. Once we know $r$, we can find the actual friction required at any lower speed.

Finding the friction at 5 m/s

1. Extract the radius from the maximum-speed condition.

   At $v_{\text{max}} = 7 \, \text{m/s}$, static friction is at its limit:

Cancel $m$ and solve for $r$:

2. Calculate the centripetal force needed at 5 m/s.

   At $v = 5 \, \text{m/s}$ on the same track (same $r = 9.8 \, \text{m}$), the required centripetal force is

3. Recognize that static friction supplies exactly this force. …

Concept & Intuition

Work done by a force on a body equals the change in its kinetic energy (work–energy theorem). Since we are given displacement as a function of time, we can differentiate to get velocity, then compute kinetic energy at the start and end of the interval. No need to find the force explicitly — that’s the elegant shortcut.

Step-by-step solution

1. Find velocity from displacement Displacement: $s = \frac{t^3}{3}$. Velocity is the time derivative:

2. Kinetic energy at any time Mass $m = 3$ kg. Kinetic energy:

3. Work done in first two seconds At $t = 0$: $K(0) = 0$ (body starts from rest). At $t = 2$ s: $K(2) = \frac{3}{2} \cdot (2)^4 = \frac{3}{2} \cdot 16 = 24$ J. …

The efficiency of any machine, including a pump, tells us how effectively it converts the energy supplied to it into useful work. In this problem, the pump consumes electrical energy (input power) to lift water against gravity, thereby increasing its potential energy (useful output power). The efficiency is simply the ratio of this useful output power to the total input power.

Here's how we calculate the pump's efficiency:

1. Identify the given quantities and the goal.

   We are given:

   • Volume of water ($V$) = $36\, \mathrm{m}^3$
   • Time taken ($t$) = $30\, \mathrm{min}$
   • Height the water is lifted ($h$) = $50\, \mathrm{m}$
   • Electric power consumed by the pump ($P_{input}$) = $40\, \mathrm{kW}$
   • Acceleration due to gravity ($g$) = $10\, \mathrm{m/s}^2$
   • Density of water ($\rho$) = $1000\, \mathrm{Kg/m}^3$

   Our goal is to find the efficiency ($\eta$) of the pump.

2. Convert all units to the SI system for consistency.

   • Time: $t = 30\, \mathrm{min} = 30 \times 60\, \mathrm{s} = 1800\, \mathrm{s}$
   • Input Power: $P_{input} = 40\, \mathrm{kW} = 40 \times 10^3\, \mathrm{W} = 40000\, \mathrm{W}$

3. Calculate the mass of water pumped.

   The mass ($m$) of the water can be found using its volume and density:

4. Calculate the potential energy gained by the water. When water is lifted to a height $h$, its gravitational potential energy ($PE$) increases. This is the useful work done by the pump.

$$PE = 36000\, \mathrm{Kg} \times 10\, \mathrm{m/s}^2 \times 50\, \mathrm{m}$$ …

The key idea is that the machine gun’s power is the rate at which it does work — here, that work goes entirely into the kinetic energy of the bullets. Since each bullet starts from rest inside the gun and leaves at 500 m/s, the energy per bullet is $\frac{1}{2} m v^2$. The power tells us how much energy is transferred per second, so if we know how many bullets are fired per second, we can solve for the mass of one bullet.

1. Find the number of bullets per second. The gun fires 360 bullets per minute.

2. Express the power in terms of bullet mass and speed. The kinetic energy of one bullet is $\frac{1}{2} m v^2$. In one second, $n$ bullets are fired, so the energy given to them per second (which is the power) is:

3. Substitute the known values. $P = 4.5 \text{ kW} = 4500 \text{ W}$, $v = 500 \text{ m/s}$, $n = 6$.

4. Simplify and solve for $m$. First, $(500)^2 = 250000$. Then $\frac{1}{2} \times 250000 = 125000$. So:

$m = \frac{4500}{750000} = \frac{45}{7500} = \frac{3}{500} = 0.006 \text{ kg}.$ …

The problem asks for the instantaneous power at a specific time, given the power at another time, for a body starting from rest under the action of a force. To solve this, we need to understand how instantaneous power, force, and velocity are related, and how these quantities change with time when a body starts from rest.

Concept and Intuition

Instantaneous power ($P$) is defined as the rate at which work is done by a force, and it can be expressed as the dot product of the force vector ($\vec{F}$) and the instantaneous velocity vector ($\vec{v}$):

If the force and velocity are in the same direction (as is the case when a force acts on a body initially at rest and causes it to move in the direction of the force), this simplifies to $P = Fv$.

When a constant force $F$ acts on a body of mass $m$, it produces a constant acceleration $a$ given by Newton's second law:

If the body starts from rest (initial velocity $u=0$), its velocity $v$ at any time $t$ is given by the kinematic equation:

Now, we can substitute these relationships into the power equation. Since $a = F/m$, we have $v = (F/m)t$.

Substituting this into $P = Fv$:

This crucial relationship shows that for a constant force acting on a body initially at rest, the instantaneous power is directly proportional to time ($P \propto t$). This is the key insight to solve the problem.

Step-by-step Derivation

1. Identify the given information:

   • Mass of the body, $m = 15\,\mathrm{kg}$.
   • Initial state: at rest ($u = 0$).
   • Instantaneous power at the end of the third second, $P_3 = 5\,\mathrm{W}$ (at $t_3 = 3\,\mathrm{s}$).
   • We need to find the instantaneous power at the end of the fourth second, $P_4$ (at $t_4 = 4\,\mathrm{s}$).

2. Assume a constant force:

   In the absence of information about how the force varies, it is standard to assume that the force acting on the body is constant. This simplifies the problem significantly and allows us to use the kinematic equations for constant acceleration.

3. Relate force, acceleration, and velocity:

   • Since the force $F$ is constant and the mass $m$ is constant, the acceleration $a = F/m$ is also constant.
   • As the body starts from rest ($u=0$), its velocity $v$ at time $t$ is given by:

4. Derive the expression for instantaneous power: The instantaneous power $P$ due to the force is given by:

Substitute the expression for $v$ from the previous step:

Now, substitute $a = F/m$:
$$P = F \left(\frac{F}{m}\right)t$$ …