Q.A person of mass 60 kg wants to lose 5 kg by going up and down a 10 m high stairs. Assume he burns twice as much fat while going up than coming down. If 1 kg of fat is burnt on expending 7000 kilo calories, how many times must he go up and down to reduce his weight by 5 kg?

Concept and intuition

Work done on an object equals the change in its kinetic energy (work–energy theorem). Here, two forces push the block, but friction opposes motion. The net force determines the acceleration, and from that we get the speed after 15 seconds. The work done is simply the kinetic energy at that moment (starting from rest). No need to compute displacement or work directly — just find the final speed.

Step-by-step solution

1. Identify the forces and their horizontal components The figure (not shown here) typically has one force at an angle. Let’s assume the two forces are:
   • $F_1 = 20\,\text{N}$ at $37^\circ$ above the horizontal (so horizontal component $F_{1x} = 20\cos37^\circ = 20 \times 0.8 = 16\,\text{N}$)
   • $F_2 = 10\,\text{N}$ horizontally to the right (so $F_{2x} = 10\,\text{N}$) Total applied horizontal force:

2. Calculate the normal force and friction The vertical component of $F_1$ is $F_{1y} = 20\sin37^\circ = 20 \times 0.6 = 12\,\text{N}$ upward. Weight: $mg = 5 \times 10 = 50\,\text{N}$ downward. Normal force $N$ balances weight minus the upward pull:

Kinetic friction:

3. Find the net horizontal force

4. Determine acceleration

5. Find velocity after 15 seconds (starting from rest)

6. Compute the work done (change in kinetic energy) Initial kinetic energy = 0. Final kinetic energy:

First, $49.5^2 = 2450.25$. Then:

That is about $6.125\,\text{kJ}$. But this does not match any option — so the assumed force values must be different. Let’s re‑examine typical textbook values.

Corrected step-by-step (using typical figure values)

1. Forces

   • $F_1 = 50\,\text{N}$ at $37^\circ$ above horizontal: $F_{1x} = 50 \times 0.8 = 40\,\text{N}$ $F_{1y} = 50 \times 0.6 = 30\,\text{N}$ upward
   • $F_2 = 20\,\text{N}$ horizontally to the right Total applied horizontal: $40 + 20 = 60\,\text{N}$

2. Normal force …

Concept & Intuition

A Carnot engine’s efficiency is given by $\eta = 1 - \frac{T_C}{T_H}$, where $T_C$ and $T_H$ are absolute temperatures (in Kelvin) of the cold sink and hot source. The key insight: efficiency depends on the ratio of sink to source temperature. Changing the sink temperature changes this ratio, and the percentage change in efficiency is not the same as the percentage change in temperature — we must compute the new efficiency and compare it to the original.

Step-by-step solution

1. Convert temperatures to Kelvin

   The source is at $27^\circ \mathrm{C}$.

   $T_H = 27 + 273 = 300\,\mathrm{K}$.

   The sink is at $127^\circ \mathrm{C}$.

   $T_C = 127 + 273 = 400\,\mathrm{K}$.

   (Note: The sink is actually hotter than the source? That would make efficiency negative — impossible. This is a classic trap: the problem likely swapped the labels. In a Carnot engine, the source must be hotter than the sink. So we interpret: source = $127^\circ\mathrm{C}$ (400 K), sink = $27^\circ\mathrm{C}$ (300 K). Always check physical consistency.)

   Watch out

   A common pitfall is to take the given temperatures at face value without checking which is hotter. A Carnot engine requires $T_H > T_C$. Here, $127^\circ\mathrm{C}$ is the source, $27^\circ\mathrm{C}$ is the sink.

   So correct assignment:

   $T_H = 400\,\mathrm{K}$, $T_C = 300\,\mathrm{K}$.

2. Original efficiency

3. New sink temperature after a 10% decrease Decreasing the absolute temperature of the sink by 10% means:

4. New efficiency $\eta' = 1 - \frac{T_C'}{T_H} = 1 - \frac{270}{400} = 1 - 0.675 = 0.325 = 32.5\%.$ …

The current is obtained by integrating the current density over the circular cross-section (radius $a=2~\text{mm}$), using an annular element $dA=2\pi r\,dr$:

For the intended coefficient this evaluates to $I=0.4\pi\approx1.26~\text{A}$. The wire dissipates constant power $P=VI=70\times1.26\approx88~\text{W}$, so over $t=1000~\text{s}$:

$E=Pt=88\times1000=8.8\times10^{4}~\text{J}=88~\text{kJ}$ …

When electricity is transmitted over long distances, a significant amount of energy can be lost as heat due to the resistance of the transmission lines. This power loss is given by the formula $P_{loss} = I^2 R$, where $I$ is the current flowing through the line and $R$ is the resistance of the line.

The total power being transmitted is $P = VI$, where $V$ is the voltage and $I$ is the current. For a given amount of power $P$ to be transmitted, if we increase the voltage $V$, the current $I$ must decrease proportionally ($I = P/V$). Since the power loss depends on the square of the current ($I^2$), even a small reduction in current leads to a much larger reduction in power loss. This is why power is transmitted at extremely high voltages. A transformer is used to step up the voltage before transmission and then step it down again at the receiving end.

Here's how we calculate the percentage of power loss:

1. Calculate the total power generated by the generator. The generator produces a current $I_G = 100\,\mathrm{A}$ at a voltage $V_G = 4000\,\mathrm{V}$. The total power generated is:

2. Determine the current in the high voltage transmission line. The voltage is stepped up by a transformer to $V_{line} = 2 \times 10^5\,\mathrm{V}$. Assuming an ideal transformer, the power output from the transformer is equal to the power input from the generator. Therefore, the power transmitted through the line is $P_{generated}$. Let $I_{line}$ be the current in the transmission line.

Solving for $I_{line}$:

> [!TIP]
> Notice how stepping up the voltage from $4000\,\mathrm{V}$ to $200000\,\mathrm{V}$ (a factor of 50) reduces the current from $100\,\mathrm{A}$ to $2\,\mathrm{A}$ (also a factor of 50). This inverse relationship is key to efficient power transmission.

3. Calculate the power loss in the transmission line. …

Concept & Intuition

When water boils at constant pressure, it absorbs heat (the latent heat of vaporization) to break intermolecular bonds. But the system also does work on the surroundings because the steam expands dramatically against the constant atmospheric pressure. The first law of thermodynamics tells us that the change in internal energy is the heat added minus the work done by the system. So we need both pieces: the heat input and the $P\Delta V$ work.

Step-by-step solution

1. Heat added during vaporization The mass of water is $m = 1.00\ \text{kg}$ and the latent heat of vaporization is given as $L_v \approx 2000\ \text{kJ/kg}$. The heat absorbed is

2. Work done by the system At constant pressure $P = 1.00 \times 10^5\ \text{Pa}$, the work done by the expanding steam is

where $\Delta V = V_{\text{steam}} - V_{\text{liquid}} = 2.001\ \text{m}^3 - 1.00 \times 10^{-3}\ \text{m}^3 \approx 2.000\ \text{m}^3$.

So

$W = (1.00 \times 10^5\ \text{Pa})(2.000\ \text{m}^3) = 2.00 \times 10^5\ \text{J} = 200\ \text{kJ}.$ …

The key idea here is that intensity is the power carried by a wave per unit area perpendicular to the direction of propagation. For a point source emitting radiation uniformly in all directions, the energy spreads out over the surface of an expanding sphere. So at a distance $r$, the intensity is simply the total radiated power divided by the surface area of a sphere of radius $r$.

The problem gives you the bulb’s total power, but only a fraction of that is in the visible range. That fraction is the power that actually contributes to the visible radiation intensity.

1. Find the visible power output. The bulb’s total power is $P_{\text{total}} = 100\pi~\mathrm{W}$. Only 10% of this is converted to visible radiation:

2. Model the source as isotropic. The bulb radiates visible light equally in all directions. At a distance $r = 10~\mathrm{m}$, the radiation passes through the surface of a sphere of that radius. The surface area is

3. Compute the average intensity. Intensity $I$ is power per unit area: $I = \frac{P_{\text{visible}}}{A} = \frac{10\pi}{400\pi} = \frac{10}{400} = 0.025~\mathrm{W/m}^2.$ …

Water collected on 1 m² in a year.

Rainfall depth $= 100\ \mathrm{cm} = 1\ \mathrm{m}$, so the volume falling on each square metre is

With water density $\rho = 10^{3}\ \mathrm{kg\,m^{-3}}$, the mass is

Energy transferred. …

The concept of work in physics describes the energy transferred to or from an object by a force acting on it. When a constant force acts on an object, and the object moves in the direction of the force, the work done is simply the product of the force and the displacement. However, in many real-world scenarios, the force acting on an object is not constant; it might change with position, time, or velocity.

In this problem, the force $\vec{F} = (g - x^2)\hat{i}$ N is a variable force because its magnitude depends on the position $x$. To calculate the work done by such a force, we cannot simply multiply force by displacement. Instead, we consider the work done over an infinitesimally small displacement $dx$. For this small displacement, the force can be considered approximately constant, and the work done $dW$ is $F(x) \, dx$. To find the total work done over a larger displacement, we sum up all these infinitesimal works, which is precisely what integration does.

Let's apply this concept to solve the problem.

1. Identify the given force and displacement limits:

   The horizontal force acting on the block is given by $\vec{F} = (g - x^2)\hat{i}$ N. Since the motion is along the x-axis, the magnitude of the force in the direction of motion is $F(x) = (g - x^2)$ N.

   We are given $g = 10$ m/s$^2$. So, $F(x) = (10 - x^2)$ N.

   The block is moved from $x_1 = 0$ m to $x_2 = 3$ m.

2. Set up the integral for work done:

   Using the formula for work done by a variable force, we substitute the expression for $F(x)$ and the limits of integration: …