Question

The minimum force required to stop a body of mass 4 kg moving along a straight line with a velocity of 54 kmph in a distance of 9 m is

• A. 75 N
• B. 100 N
• C. 50 N
• D. 25 N

Hint:

When solving physics problems, always ensure all quantities are in consistent units (preferably SI units) before applying formulas. For stopping problems, remember that the final velocity is zero and the acceleration will be negative (deceleration).

Answer 1

The Correct Option is C

Step 1: Identify the given quantities and the unknown. \begin{itemize} \item Mass of the body, $m = 4 \text{ kg}$ \item Initial velocity, $u = 54 \text{ kmph}$ \item Final velocity, $v = 0 \text{ m/s}$ (since the body stops) \item Distance, $s = 9 \text{ m}$ \item Minimum force, $F = ?$ \end{itemize} Step 2: Convert units to SI units.
The velocity is given in kmph, which needs to be converted to m/s. $u = 54 \text{ kmph} = 54 \times \frac{1000 \text{ m}}{3600 \text{ s}} = 54 \times \frac{5}{18} \text{ m/s} = 3 \times 5 \text{ m/s} = 15 \text{ m/s}$ Step 3: Calculate the acceleration (deceleration) using kinematics.
Since the force is applied to stop the body, the acceleration will be negative (deceleration). We can use the third equation of motion: $v^2 = u^2 + 2as$ Where:
$v = 0 \text{ m/s}$
$u = 15 \text{ m/s}$
$s = 9 \text{ m}$
$0^2 = (15)^2 + 2 \times a \times 9$
$0 = 225 + 18a$
$18a = -225$
$a = -\frac{225}{18} \text{ m/s}^2$
$a = -\frac{25}{2} \text{ m/s}^2 = -12.5 \text{ m/s}^2$
The negative sign indicates deceleration. The magnitude of acceleration is $12.5 \text{ m/s}^2$.
Step 4: Calculate the minimum force using Newton's second law.
Newton's second law states $F = ma$. The minimum force required to stop the body is the force that causes this calculated deceleration.
$F = m |a|$ (magnitude of force)
$F = 4 \text{ kg} \times 12.5 \text{ m/s}^2$
$F = 50 \text{ N}$
The final answer is $\boxed{\text{50 N}}$.