Question

A 1,000 kg car moving at a speed of $20\frac{m}{s}$ is brought to rest in a distance of 50 m, 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, $u=20\frac{m}{s}$,
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
$\Rightarrow F=1000\times (-4)$
$=4000\text{Newtons}$
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.