Car on ramp Answer all parts, check units and only answer if you know how to do
ID: 1466762 • Letter: C
Question
Car on ramp
Answer all parts, check units and only answer if you know how to do it
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)? A 2,282-kg car is moving up a road with a slope (grade) of 10% at a constant speed of 12 m/s What is the direction and magnitude of the frictional force? (define positive in the forward direction, i.e., up the slope)? A 1,712-kg car is moving up a road with a slope (grade) of 38% while slowing down at a rate of 1.5 m/s^2. What is the direction and magnitude of the frictional force? (define positive in the forward direction, i.e.. up the slope)? A 1,967-kg car is moving up a road with a slope (grade) of 11% while slowing down at a rate of 3.3 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
a) slope = tan(theta) = 0.23
theta = 12.95 deg
The component of gravity acceleration acting parallel to the slope is
a = gsin(theta)
a = 9.81sin 12.95
a = 2.19 m/s^2
so friction from the road on the tires must provide the additional acceleration of
a = 1.9 - 2.19 = -0.29 m/s^2
so F = ma
F = 1729*-0.29
F = - 501.41 N downslope
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2. theta = arctan(0.10) = 84.28 degrees
friction = mgsion(theta)
f = 2282*9.8*sin84.28
f = 22252.24 N up along the slope
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3. a) slope = tan(theta) = 0.38
theta = 20.806 deg
The component of gravity acceleration acting parallel to the slope is
a = gsin(theta)
a = 9.81sin 20.806
a = 3.48 m/s^2
so friction from the road on the tires must provide the additional acceleration of
a = 1.5 + 3.48 = 4.98 m/s^2
so F = ma
F = 1712 * 4.98
F = 8525.76 N N Up slope
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a) slope = tan(theta) = 0.11
theta = 6.27 deg
The component of gravity acceleration acting parallel to the slope is
a = gsin(theta)
a = 9.81sin 6.27
a = 1.07 m/s^2
so friction from the road on the tires must provide the additional acceleration of
a = 3.3 +1.07 = 4.37 m/s^2
so F = ma
F = 1967 * 4.37
F = 8595.79 N Up slope