TP 7.22

Given:

V_i = 18.0 m/s

V_f = 15.0 m/s

m = 1300 kg

Dx = 30.0 m

Find :

(a) W_net - ?

W_net = K_f - K_i = 1/2 m V_f² - 1/2 m V_i²

Since the final speed of the car is less than the initial speed, the net work done on the car is negative.

(b) F_av - ?

W_net = 1/2 m (V_f² - V_i²) = 1/2 (1300 kg) ((15.0 m/s)² - (18.0 m/s)²) = -64,350 J

on the other hand,

W_net = F_av Dx cos (180^o) = - F_av Dx

F_av = - W_net / Dx = (64 ,350 J)/(30.0 m) = 2,145 N