Question

A 1200 Kg car travels at constant speed of 32 m/s/s. If the coefficient of friction between the road and the tires is 0.45 what force is applied by the car engine to run it?

Answer 1

Answer:

Fr= 5400 N

Explanation:

Given that

m = 1200 kg

v= 32 m/s

μ = 0.45

Given that car is moving with constant speed it means that acceleration of the car is zero or we can say that total force on the car is zero.That is why ,only force needed to over come the friction force.

Fr= μ m g

Fr= 0.45 x 1200 x 10  N

Fr= 5400 N

That is why the total force should be 5400 N.

Answer 2

Answer:

Explanation:

mass, m  = 1200 kg

v = 32 m/s

coefficient of friction, μ = 0.45

As the car is moving with constant speed so the net force is zero, but the force applied by the car engine to run is the friction force.

F = μ mg = 0.45 x 1200 x 9.8 = 5292 N