THANK YOU! A 2,319-kg car is moving down a road with a slope (grade) of 11% whil

Question

THANK YOU!

A 2,319-kg car is moving down a road with a slope (grade) of 11% while speeding up at a rate of 3.8 m/s^2. What is the direction and magnitude of the frictional force?
(define positive in the forward direction, i.e., down the slope)?

A 2,047-kg car is moving up a road with a slope (grade) of 35% at a constant speed of 18 m/s. What is the direction and magnitude of the frictional force?
(define positive in the forward direction, i.e., up the slope)?

A 1,310-kg car is moving up a road with a slope (grade) of 24% while slowing down at a rate of 2.2 m/s^2. What is the direction and magnitude of the frictional force?
(define positive in the forward direction, i.e., up the slope)?

Explanation / Answer

1. Conceptually you have net force Ma = w - f; where a = 3.8 m/s^2 is the deceleration of the car, M. As the car is slowing, we have V = U - aT where U is the initial speed. So the net force -Ma showing that f > w and that means the friction force is acting opposite to the gravity force which is down. Therefore, the friction force is acting up hill. QED.

Note, we presume only the force of gravity is a driving force when the brakes are applied. That is, no engine force. Also note w and f are forces directed along the surface of the road; so they vary with the slope of the incline. In which case:

From Ma = w - f = Mg sin(11) - F cos((11))

we solve for F = M(g sin(11) - a)/cos(11)

= 2319*(9.8*sin(11)) - 3.8)/cos(11)) = -4559.60 N. ANS.

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