Question

A 1,954-kg car is moving up a road with a slope (grade) of 39% while slowing down at a rate of 3.4 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 2,149-kg car is moving up a road with a slope (grade) of 10% 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)? A 1,712-kg car is moving up a road with a slope (grade) of 24% while speeding up at a rate of 1.4 m/s^2. What is the direction and magnitude of the frictional force? (define positive in the forward direction, i.e., up the slope)?

Answer 1

1.

Slope of the road is 39%, thus the angle is,

                  theta = tan^-1 (0.39)

                           = 21.3^o

The frictional force on the car is,

         f = m[a - gsin21.3^o]

           = (1954 kg)[(3.4 m/s²) - (9.8 m/s²)sin6.3^o]

           = -312.37 N

Magnitude: 312.37 N

Direction: Negative direction (or downward direction along the slope)

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2.

The frictional force always opposes the relative motion of the object. So, the frictional force is pointed downward along the slope when it is moving up and slowing down. So, it has negative direction.

Slope of the road is 10%, thus the angle is,

                  theta = tan^-1 (0.10)

                           = 5.71^o

The frictional force on the car is,

         f = m[a - gsin5.71^o]

           = (1954 kg)[(3.8 m/s²) - (9.8 m/s²)sin5.71^o]

           = 5519.78 N

Magnitude: 5519.78 N

Direction: Negative direction (or downward direction along the slope)

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3.

Slope of the road is 24%, thus the angle is,

                  theta = tan^-1 (0.24)

                           = 13.5^o

The frictional force on the car is,

         f = m[a + gsin13.5^o]

           = (1712 kg)[(1.4 m/s²) + (9.8 m/s²)sin13.5^o]

           = 6313.45 N

Magnitude: 6313.45 N

Direction: Positive direction (or upward direction along the slope)