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A 20 kg crate is placed on a horizontal conveyor belt. The materials are such that μs = 0.41 and μk = 0.21. Draw a free-body diagram showing all the forces on the crate if the conveyer belt is speeding up to the right What is the maximum acceleration the belt can have without the crate slipping? If acceleration of the belt exceeds the value determined in part C, what is the acceleration of the crate?

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Final answer:

The maximum acceleration the belt can have without the crate slipping is equal to the maximum kinetic frictional force divided by the mass of the crate. If the acceleration of the belt exceeds this value, the crate will accelerate along with the belt.

Step-by-step explanation:

The maximum acceleration the belt can have without the crate slipping can be determined by finding the maximum frictional force between the crate and the conveyor belt. The maximum static frictional force is given by multiplying the coefficient of static friction (μs) by the normal force (N). In this case, the normal force is equal to the weight of the crate, which is 20 kg multiplied by the acceleration due to gravity (9.8 m/s²). Therefore, the maximum static frictional force is (0.41)(20 kg)(9.8 m/s²) = 80.4 N.

If the acceleration of the belt exceeds the maximum static frictional force, the crate will start to slip. In this case, the frictional force becomes the kinetic frictional force, which is given by multiplying the coefficient of kinetic friction (μk) by the normal force. Therefore, the maximum acceleration the belt can have without the crate slipping is equal to the maximum kinetic frictional force divided by the mass of the crate. In this case, the maximum kinetic frictional force is (0.21)(20 kg)(9.8 m/s²) = 40.8 N.

If the acceleration of the belt exceeds the maximum acceleration without slipping, the crate will accelerate along with the belt. The acceleration of the crate will be equal to the acceleration of the belt, since there is no friction force opposing the motion of the crate.

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