Question

A car of mass 120 kg moves over a horizontal road with a constant velocity of 40 km h -1. Then the force exerted by the engine if the coefficient of friction between the car and the road is 0.2 will be (g = 10 ms -2)

[Hint : When a body moves with a constant velocity, the applied force must be equal to the force of limiting friction]

Answer 1

Answer:

Explanation:

Let v be the maximum speed of the car which is moving around a curve of radius 40 m and it is moving with this velocity without slipping. So, frictional force = centripetal force

μmg=

R

mv

2

v

2

=μgR

v

2

=0.5×40×9.8

v

2

=196m/s

v=14m/s

μ is coefficient of friction between car and tyre.

R is the radius of curve

v is the velocity of a car