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A4.00-kg box sits on a ramp that is inclined at 30°above the horizontal.The coefficient of kinetic friction between the box and the ramp is uk=0.32.What horizontal force is required to move the box up the incline with a constant
acceleration of 3.0m/s2?

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pstnonsonjoku

From the calculation, we can see that the force that  moves the box forward is 5.7 N.

What is frictional force?

The term frictional force refers to the force that opposes motion. Given that;

ma = ukmgcosθ + F

Let F be the forward force.

Hence;

F = ma - ukmgcosθ

F = 4 * 3 - (0.32 * 4.00 * 9.8 * sin 30)

F =  5.7 N

Learn more about frictional force:https://brainly.com/question/14662717

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