Question

a car of mass 1200 kg moving at 20m/s is brought to rest over a distance of 50m. the breaking force required to stop the car
1) 2400 N
2) 4800 N
3) 7200 N
4) 1200 N​

Answer 1

Answer:

4800 N

Explanation:

F=MA

v2-u2=2as

0-20^2=2a50

a=-4

F=1200 into 4

4800 N

Answer 2

Given:

A car of mass 1200 kg moving at 20m/s is brought to rest over a distance of 50m.

To find:

Brake force ?

Calculation:

First of all, let's find the acceleration:

${v}^{2} = {u}^{2} + 2as$

$\implies \: {0}^{2} = {20}^{2} + 2a(50)$

$\implies \: 100a = - 400$

$\implies \: a = - 4 \: m {s}^{ - 2}$

Now, force applied :

$F = m \times a$

$\implies \: F = 1200 \times ( - 4)$

$\implies \: F = - 4800 \: N$

So, braking force is 4800 N.