1. A 2,372-kg car is moving down a road with a slope (grade) of 10% at a constan

Question

1. A 2,372-kg car is moving down a road with a slope (grade) of 10% at a constant speed of 16 m/s. What is the direction and magnitude of the frictional force?
(define positive in the forward direction, i.e., down the slope)?

2.A 2,319-kg car is moving down a road with a slope (grade) of 11% while slowing down 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)?

3. A 1,729-kg car is moving down a road with a slope (grade) of 23% while speeding up at a rate of 1.9 m/s^2. What is the direction and magnitude of the frictional force?
(define positive in the forward direction, i.e., down the slope)?

4. A 1,576-kg car is moving down a road with a slope (grade) of 11% while speeding up at a rate of 3.6 m/s^2. What is the direction and magnitude of the frictional force?
(define positive in the forward direction, i.e., down the slope)?

Explanation / Answer

1) "at a constant speed" means that the upslope and downslope forces are in balance.
downslope force Fg = mgcos = 2372kg * 9.8m/s² * cos10º = 22892 N
Therefore the upslope force is 22892 N.

2) 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(radians(11)) - 3.8)/cos(radians(11)) = 4413.6N

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