Question

1. A 675 kg car accelerates from 54.0 km/h to come to a stop in a distance of 175 m.  

1. Determine the net work done on the car

2. Determine the force acting on the car  

Answer 1

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)

Let me know if you've any query.