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?
Answers
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.
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.
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