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A 17 kg crate is placed on a horizontal conveyor belt. The materials are such that μs = 0.49 and μk = 0.32.

What is the maximum acceleration the belt can have without the crate slipping?

User Mirta
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Final answer:

To calculate the maximum acceleration of the conveyor belt without the crate slipping, we use the normal force and the coefficient of static friction. The maximum static frictional force, when divided by the mass of the crate, gives us the maximum possible acceleration of 4.80 m/s².

Step-by-step explanation:

The question asks for the maximum acceleration of a conveyor belt without a crate slipping, given the mass of the crate and the coefficients of static (μs) and kinetic (μk) friction between the crate and the belt.

To solve this, we first calculate the normal force (N) exerted by the crate, which is equal to its weight due to gravity (mg, where m is mass and g is the acceleration due to gravity). We then use the coefficient of static friction to find the maximum static frictional force (fs(max)).

Since friction is what keeps the crate from slipping, the maximum static frictional force is also the maximum force the conveyor belt can exert on the crate without it slipping. This force equals the mass of the crate times its maximum acceleration (amax).

First, calculate the normal force: N = mg = (17 kg)(9.8 m/s²) = 166.6 N. Next, calculate fs(max) using μs: fs(max) = μsN = (0.49)(166.6 N) = 81.63 N. Finally, solve for the maximum acceleration, amax = fs(max) / m = 81.63 N / 17 kg ≈ 4.80 m/s².

User Jonathan Burley
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