A worker pulls a 50.0 kg box up a 30 degree plane as shown in the diagram below.

Question

A worker pulls a 50.0 kg box up a 30 degree plane as shown in the diagram below. The tension in the rope is 445 N, parallel to the plane, and the box moves with a uniform velocity of 2.00 m s^-1 along the plane. At what rate is the worker doing work? What is the gain in gravitational potential energy of the box after the worker has pulled it m along the plans? How much work was done by the worker during this time? Explain the difference in the values that you have obtained for parts (c) and (b). The rope breaks and the box slows to a stop and then starts to slide down the plane. Calculate the acceleration of the box just after the rope breaks.

Explanation / Answer

a) rate at which worker doing work = F * v * cos(theta)

                                                         = 445 * 2 * cos0

                                                          = 890 W

b) gain in gravitational potential energy = 50 * 9.8 * sin30 * 25

                                                             = 6125 J

c) work was done = 445 * 25

                              = 11125 J

d) Difference in values is due to kinetic energy of the box .

e) acceleration of box = (50 * 9.8 * sin30 - 200)/50

                                      = 0.9 m/s2

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