Question

A 50 kg bicyclist, traveling at a speed of 12 m/s, applies the brakes, slowing her speed to 3 m/s.

A) How much work did the brakes do?

B) If the net force slowing her down is 120 N, over what distance did this take place?

Answer 1

a) Work done = Net Kinetic Energy

= 1/2 x 50 kg x ((12m/s)^2 - (3m/s)^2)

= 0.5 x 50 Kg x (144 -9)(m/s)^2

= 3375 Kg (m/s)^2

b) Force = mxa

a = 120 N/50 Kg = 2.4 m/s^2

Using newtons third law of motion, we get-

V^2 - U^2 = 2 x a x S

S= (12^2-3^2)m^2/s^2/(2 x 2.4 m/s^2)

= 28.125 m