A 1050-kg car is being driven up a 6.43 ° hill. The frictional force is directed

Question

A 1050-kg car is being driven up a 6.43 ° hill. The frictional force is directed opposite to the motion of the car and has a magnitude of 501 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 312 m. What should be the magnitude of F, so that the net work done by all the forces acting on the car is 129 kJ?

Explanation / Answer

given
m = 1050 kg
teta = 6.43 deg
f = 501 N
F = ?
l = 312 m

work done by gravity = -m*g*lsin(theta)
work done by friciotn = -f*l
work done by force = Fd

hence
Fd - f*l - mg*l*sin(theta) = 129,000
Fd = 755445.423763 N

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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