Question

A scooter of mass 120 kg is moving with a uniform velocity of $108km{h}^{-1}$. The force required to stop the vehicle in 10 s is

• A. 360 N
• B. 720 N
• C. 180 N
• D. $20\times 10.8N$

Answer 1

Answer: (A)

Force, F = m $\times$ a where a = $\frac{\upsilon -u}{t}$ i.e. F = m$\left(\frac{\upsilon -u}{t}\right)$
Here m = 120 kg,
$u=108\times \frac{1000}{3600}=30m{s}^{-1}$,
$\upsilon =0$
and t = 10 s \therefore F = $\frac{120(0-30)}{10}=-360N$
The force required to stop the vehicle is 360 N.
1 km $h^{-1}=\frac{5}{18}m{s}^{-1}$