A 2,284-kg car is moving down a road with a slope (grade) of 13% at a constant s

Question

A 2,284-kg car is moving down a road with a slope (grade) of 13% at a constant speed of 11 m/s. What is the direction and magnitude of the frictional force? (define positive in the forward direction, i.e., down the slope)? A 1,718-kg car is moving up a road with a slope (grade) of 14% at a constant speed of 15 m/s. What is the direction and magnitude of the frictional force? (define positive in the forward direction, i.e., up the slope)? A 1,771-kg car is moving down a road with a slope (grade) of 11% while slowing down at a rate of 3.1 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 1,718-kg car is moving up a road with a slope (grade) of 14% at a constant speed of 15 m/s. What is the direction and magnitude of the frictional force? (define positive in the forward direction, i.e., up the slope)? I would really like to see the steps how to do this. Thanks so much in advance.

Explanation / Answer

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1. The problem statement, all variables and given/known data
A 1,970-kg car is moving down a road with a slope (grade) of 14% while slowing down 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. Relevant equations
?=?*m*g*cos?
Fn = m*g*cos?
F= m*a


3. The attempt at a solution
i know m = 1970kg
a = -3m/s^2
g=-9.8m/s^2
So i converted the grade using arctan(14/100) = ~8

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