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Asked by Jan Dupre

A force of 55 N at an angle of 32 degrees to the horizontal is applied to drag a 15kg box across a horizontal floor at a velocity of 2.5 m/s. The coefficient of friction between the box and the floor is 0.25 and the box is moved 15m across the floor. How much work is done by the force applied?
9 years ago

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Answered by Henry
Wb = M*g = 15 * 9.8 = 147 N.

Fn = 147 - 55*sin32 = 117.9 N. = Normal force.

Fk = u*Fn = 0.25 * 117.9 = 29.5 N. = Force of kinetic friction.

Work = (55*Cos32-29.5) * 15 = 257 Joules.

9 years ago

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