Problem 64 - A Complete Resource Book in Physics for JEE Main

Question

A constant retarding force of \(50 \mathrm{~N}\) is applied to a body of mass \(20 \mathrm{~kg}\) moving initially with a speed of \(15 \mathrm{~ms}^{-1}\). How long does the body take to stop? (A) \(2 \mathrm{~s}\) (B) \(4 \mathrm{~s}\) (C) \(6 \mathrm{~s}\) (D) \(8 \mathrm{~s}\)

Step-by-Step Solution

Answer

The body takes 6 seconds to stop. The correct answer is (C).

Step 1: List down the given data

The retarding force: \(F = 50\text{ N}\) The mass of the body: \(m = 20\text{ kg}\) Initial velocity: \(v_0 = 15 \text{m/s}\) Final velocity: \(v_f = 0\text{ m/s}\) (since the body stops)

Step 2: Use Newton's second law to relate force, mass, and acceleration

Newton's second law states that \(F = ma\), where F is the force applied, m is the mass of the body, and a is the acceleration of the body. As the force is a retarding force (acting in the opposite direction of the motion), we should consider the force as negative. So, the equation becomes \(-F = ma\). Now we can plug in the given values to find the acceleration: \(-50\mathrm{~N} = (20\text{ kg})a\)

Step 3: Calculate acceleration

To calculate acceleration, we'll rearrange the equation from step 2 and solve for a: \(a = -\frac{50\text{ N}}{20\text{ kg}}\) \(a = -2.5 \text{m/s}^2\)

Step 4: Use the kinematic equation to relate initial and final velocities, acceleration, and time

The kinematic equation that relates initial velocity, final velocity, acceleration, and time is: \(v_f = v_0 + at\) Plug in the values we found so far: \(0\text{ m/s} = 15\text{ m/s} - (2.5 \text{m/s}^2)t\)

Step 5: Solve for time

Rearrange the equation from step 4 and solve for t: \(t = \frac{(0\text{ m/s}-15\text{ m/s})}{-2.5 \text{m/s}^2}\) \(t = \frac{-15}{-2.5}\) \(t = 6\text{ s}\) So, the body takes 6 seconds to stop. The correct answer is (C).