A 1120-kg car is being driven up a 8.17 ° hill. The frictional force is directed

Question

A 1120-kg car is being driven up a 8.17 ° hill. The frictional force is directed opposite to the motion of the car and has a magnitude of 540 N. A force F is applied to the car by the road and propels the car forward. In addition to these two forces, two other forces act on the car: its weight W and the normal force FN directed perpendicular to the road surface. The length of the road up the hill is 337 m. What should be the magnitude of F, so that the net work done by all the forces acting on the car is 205 kJ?

Explanation / Answer

Work done is given by:

W = Fnet*d

Fnet = W/d = 205000/337

Fnet = 608.31 N

Now there are total four forces, but due to Fn there is no work, So

Fnet = F + Ff + Fg

F = Fnet - Ff - Fg

Ff = -540 N

Fg = m*g*cos A = 1120*9.81*cos 98.17 deg = -1561.40 N

So,

F = 608.31 - (-540 - 1561.40)

F = 2709.71 N

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.