A 675 kg car accelerates from 54.0 km/h to come to a stop in a distance of 175 m.
Determine the net work done on the car
Determine the force acting on the car
Given that Initial speed of car, U = 54.0 km/hr = 54.0*(5/18) = 15.0 m/s
Final speed of car, V = 0 m/s
Mass of car = 675 kg
So, Using work-energy theorem:
W = dKE = KEf - KEi
W = (1/2)*m*V^2 - (1/2)*m*U^2
W = (1/2)*675*0^2 - (1/2)*675*15.0^2
W = -75937.5 J (-ve sign means work is done on the car)
So Work-done on the car = 75937.5 J
Part B.
Work-done is also given by:
W = F.d
d = displacement of car = 175 m
F = W/d = -75937.5/175
F = -433.9 N (Again -ve sign means force is applied on the car)
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