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A mover pushes a 30.0 kg crate across a wooden floor at a constant speed of 0.75 m/s. If the coefficient of static friction for wood-on-wood is 0.20, what is the normal force exerted by the floor on the crate?

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mnhlabatsi7

Answer:

294.3 N

Explanation:

In this situation, we are told that the crate is not accelerating in the horizontal plane. But also it is not accelerating in the vertical plane. Meaning that the sum of all vertical forces add up to zero.

Fnet =  ma

Weight + Normal force = mass *  acceleration

-(30 kg * 9.81 m/s²) + Normal force = 30.0 kg * 0 m/s²

                                  Normal force = 294.3 N

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