I really need help with these!!!! Thank you so much in advance. Question 1 A 1,4

Question

I really need help with these!!!! Thank you so much in advance.

Question 1 A 1,461-kg car is moving down a road with a slope (grade) of 15% 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., down the slope)? 10 points

Question 2 A 1,342-kg car is moving down a road with a slope (grade) of 15% while slowing down at a rate of 3.9 m/s^2. What is the direction and magnitude of the frictional force? (define positive in the forward direction, i.e., down the slope)?

Explanation / Answer

part A:

tan(angle of incline)=0.15

==>angle of incline=8.53076 degrees

constant speed means net force on the car is zero.

hence friction force=weight of the car along the incline

friction will oppose the motion hence it will act in upward the slope direction

magnitude of friction force=mass*g*sin(angle of the incline)=1461*9.8*sin(8.53076)=2123.9 N

part b:
as it is moving down, friction force will act in opposite direction

angle of the incline with horizontal=arctan(0.15)=8.53076 degrees

forces acting on the car:

component of weight of the car along the slope, acting in downward direction

friction force acting in upward direction
slowing down means acceleration=-3.9 m/s^2

hence using force balance along the incline:

component of weight of the car along the slope-friction force=mass*acceleration

==>1342*9.8*sin(8.53076)-friction force=1342*(-3.9)

==>friction force=7184.71 N

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