Question

After a historically poor banana harvest, Tarzan was forced to take a job as a receiving clerk at UPS. Tarzan has to move a 45.0 kg crate from the ground onto a loading dock with a height of 1.50m. He can complete this task either by lifting the crate onton the dock or by sliding the crate up a ramp with an incline angle of 35.4 degrees. The coefficient of kinetic friction for the crate/ramp system is 0.225. How much work is required for Tarzan to simply lift the crate up onto the dock? How much work is required for Tarzan to slide the crate up the ramp to the dock (at constant speed)? What is Tarzans average power output when he lifts the crate onto the dock over a period of 1.80s? WHat is Tarzans average power output when he slides the crate up the ramp over a time period of 8.00s? How much energy is "lost" to friction when the crate is slid up the ramp?

Answer 1

Given,

m = 45 kg ; h = 1.5 m ; theta = 35.4 deg ; uk = 0.225

for simply lifting:

W1 = m g h

W1 = 45 x 9.81 x 1.5 = 662.18 J

Hence, W1 = 662.18 J

for sliding up the ramp of length L,

L = h/sin(theta) = 1.5/sin35.4 = 2.6 m

W2 = uk m g cos(theta) d + mg sin(theta)

W2 = 0.225 x 45 x 9.81 x cos35.4 x 2.6 + 45 x 9.81 x sin35.4 = 466.23 J

Hence, W2 = 466.23 J

P1 = W1/t1 = 662.18/1.8 = 367.88 W

Hence, P1 = 367.88 W

P2 = 466.23/8 = 58.28 W

P2 = 58.28 W

Wf = uk m g cos(theta) d

Wf = 0.225 x 45 x 9.81 x cos35.4 x 2.6 = 210.51 J

Wf = 210.51 J