Question

A 1450 kg car begins sliding down a 12.0 ° inclined road with a speed of 34.0 km/h.The engine is turned off, and the only forces acting on the car are a net frictional force from the road and the gravitational force. After the car has traveled 45.0 m along the road, its speed is 50.0 km/h. (a) How much is the mechanical energy of the car reduced because of the net frictional force? (b) What is the magnitude of that net frictional force?

Answer 1

Given : m = 1450 kg ; = 12° ; v₁= 34 km/h = 9.44 m/s ; d = 45 m ; v₂ = 50 km/h = 13.89 m/s

Solution :

(a) Mechanical energy of the car reduced due to net frictional force

Here , the height is given by : h = -dsin

i.e. h = -(45 m)sin(12°) = -9.356 m

Change in potential energy can be calculated as :

U = mg(h)

= (1450 kg)(9.81 m/s²)(-9.356 m)

= -133.084 kJ

Change in kinetic energy is given by:

K = (1/2)m(v₂² - v₁²)

= (1/2)(1450)(13.89² -9.44²)

= 75.268 kJ

Amount of mechanical energy reduced is given by :

E = U + K

= -133.084kJ + 75.268 kJ

= -57.816 kJ

(b) Magnitude of net frictional force

We know that , E = F×d , where F is the net frictional force

-57.816 kJ = F×(45m)

F = (-57.816 kJ)/45m = 1.28 kN