Question

A 1100-kg car sits at rest at the bottom of a 15 degree hill. The driver initiates the accelerator and the engine pushes it up a 15 degree hill with a net force of 4500 N parallel to the hill. There is an effective 0.1 coefficient of kinetic friction. Use energy concepts to calculate the speed of the car after it has moved 40 m along the hill’s surface.

Answer 1

Force applied F = 4500N

Distance moved along the hill d= 40m

Work done by force W = Fd= 4500*40 = 180000J

Friction force f = - coefficient of kinetic friction * mg cos(theta)

Mass of car m = 1100 kg

Acceleration due to gravity g = 9.80 m/s2

Angle of inclination (theta) = 150

Friction force f = - 0.1*1100*9.8*cos(15) = - 1041.27N

Work done by friction Wf= 1041.27*40=41650.72J

Work done by gravity Wg = - mgdsin(theta) = 1100*9.80*40*Sin(15) =- 111602.77J

Work done by all forces W = 180000-41650.72-111602.77 = 26746.51 J

According to work energy theoremworkdone by all force = change in kinetic energy = final kinetic energy - initial kinetic energy

But initial kinetic energy = 0

Therefore final kinetic energy = 26746.51J

(1/2)mv2= 26746. 51J

Sped of the car v = sqrt(2*26746.51/1100) = 6.97m/s