A car is moving across a level highway with a speed of 29.4 m/s. The brakes are applied and the wheels become locked as the 1260-kg car skids to a stop. The braking distance is 135 meters.

Question

a.) What is the initial kinetic energy of the car?

b.) What is the final final energy of the car?

c.) How much work was done by the brakes to stop the car? 

d.) Determine the magnitude (enter your answer as a positive answer) of the constant braking force acting upon the car.

Answer 1

Hello Charlotte;

1) KE = 1/2mV²

0.5 x 1260 x 29.4² = 544,546.8 ≈ 545 KJ

2) We can find the KE if know the final velocity. Since, it stops eventually, its final energy is 0.

E_final = 0

3) W(brakes) = ΔE = E_f - E_i = 0 - 544,546.8 = -544,546.8J ≈ 545 KJ( Negative means the energy is lost to surroundings)

3) Let's find, first, the acceleration using the kinematics formula

(V_final)2 = (V_o)2 + 2ax

0 = 29.4² + 2 x a x 135 ==> a = -3.2 m/s²

F_(brake) = ma (Newton's 2^nd Law)

F_(brake) = 1260 x |-3.2 | = 4032 N ≅ 4KN (magnitude)

Alternatively; since the work on the car is done by the braking force;

E = F x d = 544,546.8 = F_(braking) x 135 = 4033.68 ≅ 4kN

I hope I could be of help. Please let me know, Charlotte, if you need clarification.