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A constant force of 10N is applied to a 2−kg crate on a rough surface that is sitting on it. The crate undergoes a frictional force against the force that moves it over the surface. (a) Assuming the coefficient of friction is \mu_k=0.24μ k​ =0.24, find the magnitude of the friction force opposes the motion?(b) What is the net force on the

WITH EXPLANATION PLEASE.

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Deriyan191293idn

Answer:

Explanation:

The kinetic friction between crate and the surface can be calculated as follows: [tex]f_{k}=\mu_{k}N=\mu_{k}mg=(0.24)(2)(10)=4.8 N[/tex]

Assume that the direction of the 10 N constant force is on the right, the kinetic friction will go the left, then:

[tex]\sum F=F_{net}=F-f_{k}=10-4.8=5.2 N[/tex]

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