1. A 1,224-kg car is moving up a road with a slope (grade) of 24% while slowing

Question

1.

A 1,224-kg car is moving up a road with a slope (grade) of 24% while slowing down at a rate of 1.4 m/s^2. What is the direction and magnitude of the frictional force?
(define positive in the forward direction, i.e., up the slope)?

2.

A 2,254-kg car is moving down a road with a slope (grade) of 30% while speeding up at a rate of 1.1 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 1,224-kg car is moving up a road with a slope (grade) of 24% while slowing down at a rate of 1.4 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

Note that

theta = Arctan (grade)

thus,

theta = 13.496 degrees

Thus, summing forces,

m g sin 13.496 + Ff = m a

as a = -1.4 m/s^2, then

Ff = -m g sin 13.496 + m a = -4513 N, up the slope [ANSWER]

***********************

Note that

theta = Arctan (grade)

thus,

theta = 16.7 degrees

Thus, summing forces,

m g sin 16.7 + Ff = m a

as a = 1.1 m/s^2, then

Ff = -m g sin 16.7 + m a = -3868 N, up the slope [ANSWER]

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