Question

A $1000kg$ 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.

• A. (-) 4000 N, More
• B. 3000 N, Less
• C. 2000 N, more
• D. 1000 N, more

Answer 1

The correct option is A

(-) 4000 N, More

For calculating force, we need to calculate deceleration first. Now,
$v^2-u^2=2aS$
Here, $u=20m/s,S=50m$ and $v=0$. Therefore,
$0-{20}^2=2\times a\times 50$
$\Rightarrow a=-4m/s^2$

From Newton's second law,
$F=ma$
$\Rightarrow F=1000\times (-4)=-4000$ Newtons
Hence, the net force is $-4000$ Newtons. The negative sign indicates that the force is acting opposite to the direction of initial velocity and causes deceleration. This net force is actually the frictional force acting between the ground and the tyre of the car.