Work, Energy and Power — NCERT Solutions

The work done by a force is given by $W = F \cdot d \cos\theta$, where $\theta$ is the angle between the force and displacement.

(a) Positive.
The man applies force upward (via rope) and the bucket moves upward. The force and displacement are in the same direction ($\theta = 0°$), so work done is positive.

(b) Negative.
Gravitational force acts downward but the bucket moves upward. The force and displacement are in opposite directions ($\theta = 180°$), so work done by gravity is negative.

(c) Negative.
Friction on a body sliding down an inclined plane acts up the plane (opposing motion), while displacement is down the plane. Since $\theta = 180°$, work done by friction is negative.

(d) Positive.
The applied force is in the direction of motion (horizontal). Even though friction also acts, the applied force itself does positive work ($\theta = 0°$).

(e) Negative.
The resistive force of air opposes the motion of the pendulum at every point. Since force and displacement are always in opposite directions ($\theta = 180°$), work done by air resistance is negative.

Given:
Mass $m = 2\,\mathrm{kg}$, Applied force $F = 7\,\mathrm{N}$, $\mu_k = 0.1$, $g = 10\,\mathrm{m\,s^{-2}}$, initial velocity $u = 0$, time $t = 10\,\mathrm{s}$.

Step 1: Find friction force.
$$f = \mu_k\, m g = 0.1 \times 2 \times 10 = 2\,\mathrm{N}$$

Step 2: Find net force and acceleration.
$$F_{\text{net}} = F - f = 7 - 2 = 5\,\mathrm{N}$$
$$a = \frac{F_{\text{net}}}{m} = \frac{5}{2} = 2.5\,\mathrm{m\,s^{-2}}$$

Step 3: Find displacement in 10 s.
$$s = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2}\times 2.5 \times (10)^2 = 125\,\mathrm{m}$$

(a) Work done by applied force:
$$W_{\text{applied}} = F \times s = 7 \times 125 = 875\,\mathrm{J}$$

(b) Work done by friction:
$$W_{\text{friction}} = -f \times s = -2 \times 125 = -250\,\mathrm{J}$$
(Negative because friction opposes displacement.)

(c) Work done by net force:
$$W_{\text{net}} = F_{\text{net}} \times s = 5 \times 125 = 625\,\mathrm{J}$$

(d) Change in kinetic energy:
Final velocity: $v = u + at = 0 + 2.5 \times 10 = 25\,\mathrm{m\,s^{-1}}$
$$\Delta K = \frac{1}{2}mv^2 - 0 = \frac{1}{2}\times 2 \times (25)^2 = 625\,\mathrm{J}$$

Interpretation: $\Delta K = W_{\text{net}} = 625\,\mathrm{J}$, which verifies the Work-Energy Theorem. The work done by the applied force (875 J) goes partly into kinetic energy (625 J) and partly is lost to friction (250 J).

Concept: The kinetic energy $K = E - V(x)$. Since $K \geq 0$, the particle can only exist in regions where $V(x) \leq E$ (total energy). Regions where $V(x) > E$ are classically forbidden.

(a) Fig. (a) — V(x) = constant for $x > 0$, rises sharply at $x = 0$:
The total energy $E$ is marked above the flat potential region. The particle can be found everywhere in the region where $V(x) \leq E$. Since the potential is a finite constant and $E$ is above it for all $x > 0$, no region is forbidden (the particle can be found everywhere to the right). The minimum total energy equals the value of $V(x)$ (the constant value of the potential).

(b) Fig. (b) — V(x) is a finite potential well:
The total energy $E$ is below the potential on both sides (outside the well). The particle cannot be found in the regions outside the well where $V(x) > E$. The minimum total energy is the value of $V(x)$ at the bottom of the well.

(c) Fig. (c) — V(x) has a potential barrier in the middle:
The total energy $E$ is below the peak of the barrier. The particle cannot be found in the region of the barrier where $V(x) > E$. The particle is confined to the region on one side of the barrier. The minimum total energy is the value of $V(x)$ at the lowest point.

(d) Fig. (d) — V(x) is a parabolic potential (like SHM):
The total energy $E$ is marked. The particle cannot be found in the regions where $V(x) > E$, i.e., beyond the turning points $x_1$ and $x_2$ where $V(x) = E$. The minimum total energy is the value of $V(x)$ at the equilibrium position (bottom of the parabola), which is zero for a harmonic oscillator.

Note: In each case, the minimum total energy the particle must have is equal to the minimum value of $V(x)$ in the accessible region, so that $K \geq 0$ everywhere.

Given:
Force constant $k = 0.5\,\mathrm{N\,m^{-1}}$, Total energy $E = 1\,\mathrm{J}$.

Concept: At the turning point, the kinetic energy is zero, so all energy is potential:
$$E = V(x) = \frac{1}{2}kx^2$$

At $x = \pm 2\,\mathrm{m}$:
$$V(\pm 2) = \frac{1}{2} \times 0.5 \times (2)^2 = \frac{1}{2} \times 0.5 \times 4 = 1\,\mathrm{J}$$

Since $V(\pm 2) = E = 1\,\mathrm{J}$, the kinetic energy at $x = \pm 2\,\mathrm{m}$ is:
$$K = E - V = 1 - 1 = 0\,\mathrm{J}$$

Since the kinetic energy becomes zero at $x = \pm 2\,\mathrm{m}$, the particle momentarily comes to rest at these points. For $|x| > 2\,\mathrm{m}$, $V(x) > E$, which would require $K < 0$ — physically impossible.

Therefore, the particle must turn back at $x = \pm 2\,\mathrm{m}$. $\hspace{2cm}\blacksquare$

(a) The heat energy required for burning the rocket casing is obtained at the expense of the rocket's kinetic energy (and hence its fuel/internal energy). The rocket is moving at high speed through the atmosphere; friction between the casing and air converts the rocket's kinetic energy into heat. The atmosphere does not supply this energy.

(b) Gravity is a conservative force. For any conservative force, the work done over a closed path is always zero. Since the comet returns to its starting point after completing one full orbit, the total displacement over the closed orbit is zero in terms of potential energy change: $W = -(\Delta V) = -(V_f - V_i) = 0$ (since $V_f = V_i$ for a complete orbit). Hence, the work done by the gravitational force over every complete orbit is zero.

(c) For a satellite in a circular orbit at radius $r$:

• Total mechanical energy: $E = -\frac{GMm}{2r}$ (negative)
• Kinetic energy: $K = \frac{GMm}{2r}$
• Potential energy: $V = -\frac{GMm}{r}$

As the satellite loses energy due to atmospheric drag, $E$ becomes more negative (decreases). This means $r$ decreases. As $r$ decreases, $K = \frac{GMm}{2r}$ increases — so the satellite speeds up. The decrease in potential energy is twice the increase in kinetic energy; the net energy lost goes to overcoming atmospheric resistance. Thus, the satellite paradoxically speeds up even as it loses total mechanical energy.

(d) Case (i): The man carries the mass horizontally. The gravitational force on the mass acts downward, but displacement is horizontal. Work done against gravity = 0. The man does no work against gravity (though he expends biological energy).

Case (ii): The man pulls the rope over a pulley, lifting the 15 kg mass. The mass moves upward. Work done against gravity:
$$W = mgh = 15 \times 10 \times 2 = 300\,\mathrm{J}$$

The work done in case (ii) is greater (300 J vs. 0 J against gravity).

(a) Decreases.
For a conservative force, $W = -\Delta V$. If the force does positive work ($W > 0$), then $\Delta V < 0$, meaning potential energy decreases.

(b) Kinetic energy.
Friction is a non-conservative force that opposes motion. Work done against friction directly reduces the speed of the body, hence its kinetic energy decreases. (Potential energy is associated only with conservative forces.)

(c) External force.
By Newton's Third Law, internal forces cancel in pairs. The rate of change of total momentum of a system equals the net external force on the system:
$$\frac{d\mathbf{P}}{dt} = \mathbf{F}_{\text{external}}$$

(d) Total linear momentum.
In an inelastic collision:

• Total kinetic energy is not conserved (some is lost to heat/deformation).
• Total linear momentum is conserved (no external forces).
• Total energy is conserved (but some KE converts to other forms, so total mechanical energy changes).

The correct answer is total linear momentum.

(a) False.
In an elastic collision, the total momentum and total kinetic energy of the system are conserved, but the momentum and kinetic energy of each individual body generally change. The bodies exchange momentum and energy during the collision.

(b) False.
Total energy is conserved only when there are no external forces doing work on the system. If external forces act on the system (e.g., an external agent does work), the total energy of the system changes. The law of conservation of energy applies to an isolated system (no external forces).

(c) False.
Work done over a closed loop is zero only for conservative forces (e.g., gravity, spring force). For non-conservative forces like friction, the work done over a closed path is not zero — it is negative (energy is dissipated).

(d) True.
In an inelastic collision, kinetic energy is not conserved. Some kinetic energy is converted into heat, sound, or deformation energy. Therefore, the final kinetic energy is always less than the initial kinetic energy. (In a perfectly inelastic collision, the loss is maximum.)

(a) No.
During the short time of contact in an elastic collision, the balls are deformed and the kinetic energy is temporarily converted into elastic potential energy (deformation energy). The total kinetic energy is not conserved at every instant during the collision. It is only conserved before and after the collision (i.e., the total KE returns to its original value once the balls separate).

(b) Yes.
Total linear momentum is conserved at every instant during the collision, including during the contact period. This follows from Newton's Third Law: the forces the balls exert on each other are equal and opposite, so the net internal force is zero and total momentum does not change.

(c)

• Kinetic energy (inelastic): Not conserved during contact (converted to deformation/heat), and also not fully recovered after the collision. The final KE is less than the initial KE.
• Linear momentum (inelastic): Yes, conserved at every instant during the collision, just as in the elastic case, because Newton's Third Law still applies.

(d) Elastic.
If the potential energy depends only on the separation between centres, it is a conservative force. During the collision, kinetic energy converts to this potential energy and then fully converts back to kinetic energy as the balls separate. Since no energy is permanently lost, the collision is elastic.

Correct option: (ii) $t$

Reasoning:
For a body starting from rest with constant acceleration $a$:
$$v = at$$
The net force is $F = ma$ (constant).
Power delivered:
$$P = F \cdot v = ma \cdot at = ma^2 t$$
Since $m$, $a$ are constants:
$$\boxed{P \propto t}$$

Correct option: (iii) $t^{3/2}$

Reasoning:
Constant power $P = Fv = mav =$ constant.
Since $P =$ constant:
$$P = mv\frac{dv}{dt} = \text{const}$$
$$v\,dv = \frac{P}{m}\,dt$$
Integrating:
$$\frac{v^2}{2} = \frac{P}{m}t \implies v = \sqrt{\frac{2P}{m}}\cdot t^{1/2}$$
Now, displacement:
$$s = \int_0^t v\,dt = \sqrt{\frac{2P}{m}}\int_0^t t^{1/2}\,dt = \sqrt{\frac{2P}{m}}\cdot\frac{2}{3}t^{3/2}$$
$$\boxed{s \propto t^{3/2}}$$

Given:
$$\mathbf{F} = -\hat{i} + 2\hat{j} + 3\hat{k}\,\mathrm{N}$$
Displacement along $z$-axis: $\mathbf{d} = 4\hat{k}\,\mathrm{m}$

Work done:
$$W = \mathbf{F} \cdot \mathbf{d} = (-\hat{i} + 2\hat{j} + 3\hat{k})\cdot(4\hat{k})$$
$$W = (-1)(0) + (2)(0) + (3)(4) = 0 + 0 + 12$$
$$\boxed{W = 12\,\mathrm{J}}$$

Only the $z$-component of force contributes to work since displacement is along the $z$-axis.

Given:
$K_e = 10\,\mathrm{keV} = 10\times10^3\times1.60\times10^{-19} = 1.60\times10^{-15}\,\mathrm{J}$
$K_p = 100\,\mathrm{keV} = 100\times10^3\times1.60\times10^{-19} = 1.60\times10^{-14}\,\mathrm{J}$
$m_e = 9.11\times10^{-31}\,\mathrm{kg}$, $m_p = 1.67\times10^{-27}\,\mathrm{kg}$

Speed of electron:
$$K_e = \frac{1}{2}m_e v_e^2 \implies v_e = \sqrt{\frac{2K_e}{m_e}} = \sqrt{\frac{2\times1.60\times10^{-15}}{9.11\times10^{-31}}}$$
$$v_e = \sqrt{3.512\times10^{15}} = 5.93\times10^7\,\mathrm{m\,s^{-1}}$$

Speed of proton:
$$v_p = \sqrt{\frac{2K_p}{m_p}} = \sqrt{\frac{2\times1.60\times10^{-14}}{1.67\times10^{-27}}}$$
$$v_p = \sqrt{1.916\times10^{13}} = 1.38\times10^6\,\mathrm{m\,s^{-1}}$$

Comparison: $v_e = 5.93\times10^7\,\mathrm{m\,s^{-1}} \gg v_p = 1.38\times10^6\,\mathrm{m\,s^{-1}}$

The electron is faster.

Ratio of speeds:
$$\frac{v_e}{v_p} = \frac{5.93\times10^7}{1.38\times10^6} \approx 43$$

$$\boxed{\frac{v_e}{v_p} \approx 43}$$