Question

1. A 1,761-kg car is moving down a road with a slope (grade) of 12% at a constant speed of 17 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 1,892-kg car is moving down a road with a slope (grade) of 12% while slowing down at a rate of 3.7 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 1,315-kg car is moving up a road with a slope (grade) of 15% while slowing down at a rate of 4.5 m/s^2. What is the direction and magnitude of the frictional force?
(define positive in the forward direction, i.e., up the slope)?

5. A 1,521-kg car is moving up a road with a slope (grade) of 27% while speeding up at a rate of 1.7 m/s^2. What is the direction and magnitude of the frictional force?
(define positive in the forward direction, i.e., up the slope)?

Please show your work. I tried doing f=mgsin(theta) its not that. Please help!!!!

Answer 1

net force Ma = W-f

also Use V = u-at   ( acs csar decelerates)

so Ma = W-f   = Mg sin theta - F cos theta

F   = M(g sin theta -a)/cos theta is the formua in all cases

so here apply

a. F = 1762 * (9.8 sin(12rad) -17)/(cos 12 rad)

F = -46476. 43 N
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b. F   = 1892 * (9.8 sin(12rad) -3.7)/(cos 12 rad)

F = -20085.6 N

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c. F = 1729 * (9.8 sin(23rad) -1.9)/(cos 23 rad)

F = 33075 .329 N

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d. F = 1315 * (9.8 sin(15rad) 4.5)/(cos 15 rad)

F = -3241.80 N
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e. F =   1521 * (9.8 sin(27 rad) -1.7)/(cos 27 rad)

F = -39946.244 N