Question

1. A 2,218-kg car is moving down a road with a slope (grade) of 24% while speeding up at a rate of 1.3 m/s^2. What is the direction and magnitude of the frictional force?

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?

3. A 1,762-kg car is moving up a road with a slope (grade) of 11% while slowing down at a rate of 2.8 m/s^2. What is the direction and magnitude of the frictional force?

4. A 2,180-kg car is moving up a road with a slope (grade) of 13% while slowing down at a rate of 3.9 m/s^2. What is the direction and magnitude of the frictional force?

5. A 2,207-kg car is moving up a road with a slope (grade) of 23% while speeding up at a rate of 3.1 m/s^2. What is the direction and magnitude of the frictional force?

Physics

Answer 1

1) theta = arctan(0.24) = 13.5 degrees downward

Suppose friction is not there then acceleration would have been = g*sin(theta)

= 9.81*sin(13.5) = 2.289 m/sec^2

But car is sppeding up at only 1.3 m/sec^2, this means friction is in upward direction.

Magnitude of friction force = m * (g*sin(theta) - a)

= 2218*(9.81*sin(13.5) - 1.3)

= 2196 N

2) theta = arctan(0.32) = 17.74 degrees upward

Suppose friction is not there then acceleration then car will slow down

But car is maintaining constant speed this means friction is in upward direction.

Magnitude of friction force = m * g*sin(theta)

= 2073*9.81*sin(17.74)

= 6197.95 N

3) theta = arctan(0.11) = 6.27 degrees upward

Suppose friction is not there then acceleration would have been = g*sin(theta)

= 9.81*sin(6.27) = 1.0722 m/sec^2 in downward direction

But car is slowing down up at 2.8 m/sec^2 (greater than 1.0722) this means additional friction is helping to slow down. Hence friction force is in downward direction.

Magnitude of friction force = m * (a - g*sin(theta) )

= 1762*( 2.8 - 9.81*sin(6.27))

= 3045.8 N

4) theta = arctan(0.13) = 7.407 degrees upward

Suppose friction is not there then acceleration would have been = g*sin(theta)

= 9.81*sin(7.407)

= 1.264 m/sec^2 in downward direction

But car is slowing down up at 3.9 m/sec^2 (greater than 1.264) this means additional friction is helping to slow down. Hence friction force is in downward direction.

Magnitude of friction force = m * (a - g*sin(theta) )

= 2180 *( 3.9 - 9.81*sin(7.407))

= 5745.01 N

5) theta = arctan(0.23) = 12.95 degrees upward

Suppose friction is not there then acceleration would have been = g*sin(theta)

= 9.81*sin(12.95)

= 2.198 m/sec^2 in downward direction

But car is speeding up at 3.1 m/sec^2 , this means additional friction is overcoming gravitation as well as providing force for acceleration upwards. Hence friction force is in upward direction.

Magnitude of friction force = m * (a + g*sin(theta) )

= 2207 *( 3.1 + 9.81*sin(12.95))

= 11693.62 N