Asked by emma
a delivery person pulls a 20.8 kg box across the floor. the force exerted on the box is 95.6n (35 above the horizontal). the force of kinetic friction on the box has a magnitude of 75.5n. the box starts from rest. using the work-energy theorem determine the speed of the box after being dragged 0.750m.
Answers
Answered by
R_scott
work done by person ... 95.6 N * cos(35º) * .750 m
work done by friction ... 75.5 N * .750 m
the difference in the two work amounts is the kinetic energy of the box
K.E. = 1/2 * mass * speed^2
work done by friction ... 75.5 N * .750 m
the difference in the two work amounts is the kinetic energy of the box
K.E. = 1/2 * mass * speed^2
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