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3 years ago

7

If the coefficient of kinetic friction between a 29 kg crate and the floor is 0.38, what horizontal force is required to move th

e crate at a steady speed across the floor?

1 answer:

OLEGan [10]3 years ago

7 0

Let <em>F</em> be the magnitude of the force needed to keep the box sliding at constant speed, and <em>f</em> the magnitude of kinetic friction. Then by Newton's second law, the net horizontal force on the crate is

<em>F</em> - <em>f</em> = 0

so that

<em>F</em> = <em>f</em>

<em />

<em>f</em> is proportional to the magnitude of the normal force <em>n</em>,

<em>f</em> = 0.38<em>n</em>

The net vertical force is

<em>n</em> - <em>mg</em> = 0

which tells us that

<em>n</em> = <em>mg</em> = (29 kg) (9.8 m/s²) ≈ 284 N

Then the required force must have magnitude

<em>F</em> = <em>f</em> = 0.38 (284 N) ≈ 110 N

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