PLEASE EXPLAIN EVERYTHING! 1. A 1,970-kg car is moving down a road with a slope
ID: 1338817 • Letter: P
Question
PLEASE EXPLAIN EVERYTHING!
1. A 1,970-kg car is moving down a road with a slope (grade) of 14% while speeding up at a rate of 3 m/s^2. What is the direction and magnitude of the frictional force?
(define positive in the forward direction, i.e., down the slope)
2. A 2,073-kg car is moving up a road with a slope (grade) of 32% at a constant speed of 14 m/s. What is the direction and magnitude of the frictional force?
(define positive in the forward direction, i.e., up the slope)?
3.A 2,210-kg car is moving up 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., up the slope)?
Explanation / Answer
1) theta = arctan(0.14) = 7.96 degrees downward
due to friction accleration a = g*sin(theta)
a = 9.81*sin(7.96)
a = 1.358 m/sec^2
But car is sppeding up at only 3 m/sec^2
this means friction is in upward direction.
Magnitude of friction force = m * (g*sin(theta) - a)
= 1970*(1.358 -3)
= 3234 N, up the slope
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2. theta = tan^-1(0.32) = 17.74 deg
a = 9.81 * sin 17.74
a = 2.98 m/s^2
F = 2073* (14-2.98)
F = 22844.46 N
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3. theta = tan^-1(0.11) = 6.27 deg
a = 9.81 sin 6.27 = 1.072 m/s^2
F = 2210 * (3.8-1.072)
F = 6028.88 N
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