Question

A $1,000kg$ car moving at a speed of $20m{s}^{-1}$ is brought to rest in a distance of $50m$. Calculate the net force acting on the car. Take the direction of movement of car as the positive direction.

• A. $-4000N$
• B. $4000N$
• C. $2000N$
• D. $-2000N$

Answer 1

The correct option is A $-4000N$

Given:
Mass of car $m=1000kg$
Initial velocity $u=20m{s}^{-1}$
Distance travelled $s=50m$
Car is brought to rest, so final velocity $v=0m{s}^{-1}$

Let $a$ be the acceleration of the car before coming to rest.
Let $F$ be the force acting on the car.

For calculating force, we need to calculate acceleration first.
Using second equation of motion:
$v^2-u^2=2as$
$0^2-{20}^2=2\times a\times 50$
$⟹a=-4m/s^2$

From Newton's second law of motion,
$F=ma$
$F=1000\times (-4)$
$⟹F=-4000N$

Note- The negative sign indicates that the force is acting opposite to the direction of initial velocity and causes deceleration.