Question

Select the correct alternative(s).

• A. The work done by kinetic friction acting with in system on system is always negative.
• B. Work done by static friction may be positive, negative and zero.
• C. If a particle moves in a straight line with constant acceleration under a constant force then power developed by this force varies linearly with displacement.
• D. If kinetic energy of particle moving in straight line is proportional to time then magnitude of force acting on particle is inversely proportional to speed of particle.

Answer 1

Answer: (A), (B), (D)

A, B, D

Answer 2

The question asks us to identify the correct statement(s) among the given alternatives. Let's analyze each statement:

(A) The work done by kinetic friction acting within system on system is always negative.

Consider a system of two bodies in relative motion, say body 1 and body 2. Let $\vec{f}_{12}$ be the kinetic friction force exerted by body 2 on body 1, and $\vec{f}_{21}$ be the kinetic friction force exerted by body 1 on body 2. By Newton's third law, $\vec{f}_{12} = -\vec{f}_{21}$.

The total work done by these internal kinetic friction forces on the system is $W_{internal} = \vec{f}_{12} \cdot \vec{d}_1 + \vec{f}_{21} \cdot \vec{d}_2$, where $\vec{d}_1$ and $\vec{d}_2$ are the displacements of body 1 and body 2, respectively.

Substituting $\vec{f}_{21} = -\vec{f}_{12}$, we get: $W_{internal} = \vec{f}_{12} \cdot \vec{d}_1 - \vec{f}_{12} \cdot \vec{d}_2 = \vec{f}_{12} \cdot (\vec{d}_1 - \vec{d}_2) = \vec{f}_{12} \cdot \vec{d}_{rel}$, where $\vec{d}_{rel} = \vec{d}_1 - \vec{d}_2$ is the relative displacement of body 1 with respect to body 2.

Kinetic friction always opposes the relative motion between surfaces. This means the kinetic friction force $\vec{f}_{12}$ on body 1 is always opposite to the relative velocity $\vec{v}_{rel} = \vec{v}_1 - \vec{v}_2$. Since $\vec{v}_{rel}$ and $\vec{d}_{rel}$ are in the same direction, $\vec{f}_{12}$ is opposite to $\vec{d}_{rel}$.

Therefore, the angle between $\vec{f}_{12}$ and $\vec{d}_{rel}$ is $180^\circ$.

So, $\vec{f}_{12} \cdot \vec{d}_{rel} = f_{12} d_{rel} \cos(180^\circ) = -f_{12} d_{rel}$.

Since $f_{12} > 0$ and $d_{rel} \ge 0$, the work done $W_{internal}$ is always negative or zero (if no relative displacement).

Thus, statement (A) is correct.

(B) Work done by static friction may be positive, negative and zero.

• Zero Work: When a body is at rest on a surface, static friction acts but no displacement occurs, so work done is zero. For a wheel undergoing pure rolling, the point of contact with the ground is instantaneously at rest, so static friction at this point does no work.
• Positive Work: When a person walks, static friction acts on their feet in the forward direction, propelling them forward. The displacement is in the same direction as the force, so work done is positive.
• Negative Work: When a car applies brakes and the wheels are still rolling (not skidding), static friction acts on the wheels opposite to the direction of motion. The displacement is in the direction of motion, so work done is negative.

Thus, statement (B) is correct.

(C) If a particle moves in a straight line with constant acceleration under a constant force then power developed by this force varies linearly with displacement.

Let the constant force be $F$. Since the acceleration is constant, let it be $a$.

The power developed by the force is $P = Fv$, where $v$ is the instantaneous velocity.

For constant acceleration, we have the kinematic equation $v^2 = u^2 + 2as$, where $u$ is the initial velocity and $s$ is the displacement.

From this, $v = \sqrt{u^2 + 2as}$.

Substituting this into the power equation: $P = F\sqrt{u^2 + 2as}$.

This relationship is not linear with displacement $s$. It varies as the square root of a linear function of $s$.

For instance, if $u=0$, then $P = F\sqrt{2as} = (F\sqrt{2a})\sqrt{s}$, which shows $P \propto \sqrt{s}$. This is not linear.

Thus, statement (C) is incorrect.

(Note: Power varies linearly with time: $P = F(u+at) = Fu + Fat$.)

(D) If kinetic energy of particle moving in straight line is proportional to time then magnitude of force acting on particle is inversely proportional to speed of particle.

Given that kinetic energy $K$ is proportional to time $t$: $K = ct$, where $c$ is a constant.

The power developed by the net force acting on the particle is $P = \frac{dK}{dt}$.

$P = \frac{d}{dt}(ct) = c$.

So, the power developed is constant.

Also, power is related to force and velocity by $P = Fv$ (since the particle moves in a straight line, and we consider the component of force along the velocity).

Since $P = c$ (constant), we have $Fv = c$.

Therefore, the magnitude of the force $F = \frac{c}{v}$.

This shows that the magnitude of the force is inversely proportional to the speed of the particle ($F \propto \frac{1}{v}$).

Thus, statement (D) is correct.

Based on the analysis, statements (A), (B), and (D) are correct.