A $1, 000 k g$ vehicle moving with a speed of $20 m s^{- 1}$ is brought to rest in a distance of $50 m$ . Calculate the unbalanced force acting on the vehicle.

- 4000 N

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

4000 N

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1600 N

No worries! We‘ve got your back. Try BYJU‘S free classes today!

- 1600 N

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $- 4000 N$
Given,
Initial velocity(u) $= 20 m s^{- 1}$
Final velocity(v) $=$ 0
Distance moved(S) $=$ 50m
We know,
$v^{2} = u^{2} + 2 a S$
0 $= 20^{2} + 2 a \times 50$
$- 400 = 100 a$
$a = \frac{- 400}{100}$
$a = - 4 m s^{- 2}$
Now,
$F = m a$
$F = 1000 \times - 4$
$= - 4000 N$

Suggest Corrections

Similar questions

A 1,000 kg car moving at a speed of 20 m s^–1 is brought to rest in a distance of 50 m. Calculate the net force acting on the car.

A 1000 kg vehicle moving with a velocity of 20 m/s is brought to rest in a distance of 50 m. Find the acceleration. Calculate the unbalanced force acting on the vehicle

A 1,000 kg car moving at a speed of $20 m s^{- 1}$ is brought to rest at a distance of 50 m, calculate the unbalanced force acting on the car and give a reason why the actual force applied by the brakes may be slightly less than that calculated?

A 1000kg vehicle moving with a speed of 20m/s is brought to rest in a distance of 50 meters:

(i) Find the acceleration

(ii) Calculate the unbalanced force acting on the vehicle.

Q. (a) State and explain Newton's second law of motion.
(b) A 1000 kg vehicle moving with a speed of 20 m/s is brought to rest in a distance of 50 metres :
(i) Find the acceleration
(ii) Calculate the unbalanced force acting on the vehicle.

Join BYJU'S Learning Program

A 1,000kg vehicle moving with a speed of 20ms−1 is brought to rest in a distance of 50m. Calculate the unbalanced force acting on the vehicle.

The correct option is C −4000NGiven,Initial velocity(u)=20ms−1Final velocity(v)=0Distance moved(S)=50mWe know,v2=u2+2aS0=202+2a×50−400=100aa=−400100a=−4ms−2Now,F=maF=1000×−4=−4000N

A $1, 000 k g$ vehicle moving with a speed of $20 m s^{- 1}$ is brought to rest in a distance of $50 m$ . Calculate the unbalanced force acting on the vehicle.