Question

If a bike with a rider having a total mass of 63 kg brakes and reduces its velocity from 8.5 m/s to 0 m/s in 3 second. What is the magnitude of the braking force?

Answer 1

Explanation:

m = 63 kg

Vo = 8.5 m/s

Vf = 0 m/s

t = 3 s

a = (Vf - Vo)/t

a = (0 m/s - 8.5 m/s)/3s

a = (-8.5 m/s)/3s

a = -17/6 m/s²

there is a decceleration that took place.

F = m×|a|

F = 63 kg × |-17/6 m/s²|

F = 178.5 kg m/s² ANSWER

Answer 2

The magnitude of the braking force will be 178.29 N

The initial velocity of the bike is (u) = 8.5 m/s

The final velocity of the bike is (v) = 0 m/s

Time required to reduce the speed is (t) = 3 s

From the equation of one directional motion, we get,

If deceleration is = a, then,

a = (u-v)/t

∴ a = (8.5-0)/3 m/s²

⇒ a = 2.83 m/s²

Now, the total mass (m) of the bike and the rider is = 63 Kg

∴ The braking force, F = m×a

∴ F = 63 × 2.83 N

⇒ F = 178.29 N

So, the magnitude of the braking force is 178.29 N.