Question

A body of mass 4 kg is moving with a velocity 6 ms\(^{-1}\) on a smooth surface. If it is brought to rest in 4 s, the force applied is:

• A. \(3 { N}\)
• B. \(6 { N}\)
• C. \(8 { N}\)
• D. \(4 { N}\)

Hint:

Remember, the force required to bring an object to rest is equal to the negative of the change in momentum per unit time, reflecting the force's direction opposite to the object's initial motion.

Answer 1

The Correct Option is A

Step 1: Using the formula for force \( F = \frac{\Delta p}{\Delta t} \), where \( \Delta p \) is the change in momentum and \( \Delta t \) is the time. 
Since the initial velocity \( u = 6 { ms}^{-1} \) and the final velocity \( v = 0 { ms}^{-1} \), and the mass \( m = 4 { kg} \), the change in momentum \( \Delta p \) is: \[ \Delta p = m(v - u) = 4 { kg} \times (0 - 6 { ms}^{-1}) = -24 { kg ms}^{-1}. \] The negative sign indicates a decrease in momentum. The time \( \Delta t \) is 4 s, so the force applied is: \[ F = \frac{\Delta p}{\Delta t} = \frac{-24 { kg ms}^{-1}}{4 { s}} = -6 { N}. \] The negative sign indicates the force is in the opposite direction of motion. 
Since force is a vector quantity and we are asked for the magnitude: \[ |F| = 6 { N}. \]