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Q 1-10

Marks: 1.0

You push a 35.5-kg wooden box across a wooden floor at a constant

speed of 1.5 m/s. The coefficient of kinetic friction is 0.15. How large

is the force that you exert on a box?



Sagot :

The net force on the box perpendicular to the floor is

∑ F[perp] = F[normal] - mg = 0

where mg is the weight of the box. Then

F[normal] = (35.5 kg) g = 347.9 N

Since the box slides with constant speed, the net force parallel to the floor is

∑ F[para] = F[applied] - F[friction] = 0

We have

F[friction] = 0.15 F[normal] = 52.185 N

so that

F[applied] = F[friction] ≈ 52 N