A body of mass 2.5\text{ kg} moving at a speed of 4.0\text{ m s}^{-1} along a straight hor… — JAMB UTME

A body of mass $2.5\text{ kg}$ moving at a speed of $4.0\text{ m s}^{-1}$ along a straight horizontal path is brought to rest in $2.0\text{ s}$ by a constant retarding force. What is the magnitude of this retarding force in newtons?

Answer: 5 N

Answer

The magnitude of the retarding force is

5.0\text{ N}

According to Newton's second law of motion, net force is equal to the rate of change of momentum (

F = \frac{\Delta p}{\Delta t} = \frac{m(v-u)}{t}

). Substituting

m = 2.5\text{ kg}

u = 4.0\text{ m s}^{-1}

v = 0\text{ m s}^{-1}

, and

t = 2.0\text{ s}

gives

F = \frac{2.5 \times (0 - 4.0)}{2.0} = -5.0\text{ N}

. The magnitude of this force is

5.0\text{ N}

Step-by-Step Solution

Determine the initial momentum and final momentum of the body.

Initial momentum

p_i = 2.5 \times 4.0 = 10.0\text{ kg m s}^{-1}

, and final momentum

p_f = 0\text{ kg m s}^{-1}

Linear momentum is defined as the product of mass and velocity (

p = mv

Calculate the magnitude of the force applied using the impulse-momentum relationship

F = \frac{\Delta p}{\Delta t}

Magnitude of force

F = \frac{|0 - 10.0|}{2.0} = 5.0\text{ N}

Newton's second law states that the rate of change of momentum is equal to the net external force applied.

Key Concept

Newton's Second Law and Impulse-Momentum Relationship

Estimated Time:45s

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Topics

More questions from this topic