A a 3000 lb car engine develops a useful output power of 40

A) a 3,000- lb car engine develops a useful output power of 40 hp when traveling at 60 mph (88 ft/s) on a level road. Find the drag force

B) if this same car goes up a 6% grade at 60 mph, and the drag is the same, find the power of the engine required

(A) P = F.v

P = 40 x 745.7 J and v = 60 x 1609 m / 3600s = 26.8 m/s

(40 x 745.7 J) = F (26.8)

F = 1112.3 N Or 250 lbs

(B) now force due to gravity will also comes to action.

force provided by engine, F = Fdrag + mgsin6

F = 250 + (3000 sin6) = 563.6 lbf Or 2505 N

P = F.v = 2505 x 26.8 = 67134 W

P(in hp) = 67134 / 745.7 = 90 hp