A 4.00 kg block intially sits at rest on a conveyor belt, until the power is tur
ID: 2007833 • Letter: A
Question
A 4.00 kg block intially sits at rest on a conveyor belt, until the power is turned on, and the conveyor belt starts speeding up. The coeffiecients of static and kinetic friction between the block and the conveyor belt are 0.500 and 0.300, respectively.(a) Calculate the magnitude of the maximum acceleration that the conveyor belt can have without the block slipping, rounding your answer to 2 significant figures.
_____m/s^2
(b) If the acceleration exceeds the maximum value in part (a), so that the block starts slipping. What is the magnitude of the block's resulting acceleration? Round your answer to 2 significant figures.
______m/s^2
Explanation / Answer
F=MA and F=N
The weight is 4.00 kg, so the force against the belt is F=MA, where A is the acceleration from gravity (9.81m/s/s). So the normal force applied on the belt from the block is 4kg x 9.81 m/s/ = 39.24 Newtons.
The force parallel to the belt while before the block slips is F=M, or 39.24 N x 0.500 = 19.6 Newtons. Again F=MA, so 19.6 newtons / 4kg = 4.91 m/s/s. So when the block has not started to slip, the acceleration is 4.91 m/s/s.
When the block starts to slip and is reduced to .300.The force is reduced to F= 39.24 x .300 = 11.8 N. Then again F=MA, so the acceleration is 11.8 N / 4kg = 2.95
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