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A worker pushes horizontally on a 38.0 kg crate with a force of magnitude 114 N. The coefficient of static friction between the crate and the floor is 0.390. What is the value of [tex]f_{s,\text{max}}[/tex] under the circumstances?

Answer :

Final answer:

The maximum static friction or fs,max under the given conditions is 145.0 N. This was calculated using the formula for maximum static friction, fs,max = μsN, with the given mass, gravitational acceleration, and coefficient of static friction.

Explanation:

The question asks for the maximum static friction (fs,max) which is given by the formula fs,max = μsN, where μs is the coefficient of static friction and N is the normal force. The normal force (N) for a horizontal surface is the weight of the object, given by N = mg where m is the mass of the object (38.0 kg in this case) and g is the acceleration due to gravity (9.80 m/s²).

So, N = (38.0 kg)(9.80 m/s²) = 372.4 N. Therefore, fs,max = μsN = (0.390)(372.4 N) = 145.0 N. So, the maximum static friction (fs,max) under these circumstances is 145.0 N.

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