Question

A 100 kg vehicle moving with a speed of 10 m/s is brought to rest in a distance of 50 metres. The unbalanced force acting on the vehicle will be: ​

Answer 1

Given:-

• Mass of vehicles (m) = 100kg

• Initial velocity (u) = 10m/s

• Final velocity (v) = 0m/s

• Distance covered (s) = 50m

To Find:-

• Unbalanced Force acting on the vehicles.

Solution:-

Firstly we have to calculate the acceleration of the vehicles . So by using 3rd equation of motion

• v² = u² +2as

Substitute the value we get

→ 0² = 10² + 2×a×50

→ 0 = 100 + 100×a

→ a = 100/100

→ a = 1m/s²

Now we know that Force is the product of mass and acceleration

→ F = m×a

Substitute the value we get

→ F = 100×1

→ F = 100N

∴ The unbalanced force acting on the vehicles is 100 Newton.

Answer 2

Answer:

Initial velocity(u)=20ms

−1

Final velocity(v)=0

Distance moved(S)=50m

We know,

v

2

=u

2

+2aS

0=20

2

+2a×50

−400=100a

a=

100

−400

a=−4ms

−2

Now,

F=ma

F=1000×−4

=−400N