A mover pushes a 25-kg box up a 3.5-m ramp inclined 37° upward from the horizont

Question

A mover pushes a 25-kg box up a 3.5-m ramp inclined 37° upward from the horizontal.

a. If the worker must at minimum apply 28 N parallel to the ramp, what is the
coefficient of kinetic friction between the ramp and the box?

b. How much work does the worker do in pushing the box up the ramp?

c. How much work does the force of friction while the box is moved up the
ramp?

d. How much work does the force of gravity do while the box is moved up the
ramp?

e. How much work does the normal force do while the box is moved up the
ramp?

Explanation / Answer

Mass of the box, m = 25 kg Length of the plane, L = 3.5 m Angle of inclination, = 37o Minimum force applied, F = 28 N (a) Frictional force, fk = k m g cos For uniform motion, Frictional force = External force k m g cos = F k * 25 * 9.8 * cos 37o = 28 Coefficient of kinetic friction, k = 0.143 (b) Work done by the worker, W = F * L                                               = 28 * 3.5                                              = 98 J (c) Work done by the friction, W = - k m g cos * L                                               = - 28 * 3.5                                               = - 98 J (d) Work done by the gravity, W = - mg * (L sin)                                               = - 25 * 9.8 * 3.5 * sin 37o                                               = - 516 J (d) Work done by the normal force, W = mg * (L sin)     = 25 * 9.8 * 3.5 * sin 37o     = 516 J

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