Ancient builders pushing blocks around...the return: On an earlier assignment we
ID: 1793852 • Letter: A
Question
Ancient builders pushing blocks around...the return: On an earlier assignment we considered work crews on an ancient building project moving big stone blocks around with the help of hanging masses. Let’s look at that a bit more, using the new methods available to us. Again, let’s think about the situation shown in Figure 1(a). The stone block has a mass of M = 4.50 × 103 kg (four-and-a-half tonnes!). The coefficient of static friction between the block and the ground is 0.600. Suppose we hang masses adding up to m = 2.20 × 103 kg from the end of the rope. There is a work crew pushing the block to the right. The coefficient of kinetic friction between the block and the ground is 0.550. The block is sliding to the right. (a) Over some time the block moves a distance of 5.0 m. Calculate the work done on the block by friction. Calculate the work done on the block by the tension in the rope. (b) If the block is moving at constant speed then what must the work done by the crew be? Explain. (c) Use your answer to part 1b to calculate the force exerted by the work crew. You can look back at assignment #6 to check that your answer is correct, but as usual marks are for the process, not so much for the answer. (d) Now suppose the block is on a ramp that is angled up by 10 and the crew is again pushing the block so it goes up the ramp at a constant speed. Consider the system to be the block, the ramp and the Earth. Draw an energy bar chart for the situation showing energy at some time and some later time when the block is farther up the ramp. (e) Calculate the work done by friction and by tension again if the block goes 5.0 m up the ramp. (f) Calculate the work done by the crew again and use it to calculate the force that the work crew had to exert. Again, you will be able to check your answer by comparing with Assignment #6.
Explanation / Answer
The normal force exerted by the earth surface will be
N = M2g = 4.5E4 N
hence the kinetic friction will be F = uN = .55 * N = 2.475E4 N
Hence work done by the friction will be W = F*d = 123 kJ
Tension in the rope will be equal to T = M1g = 2.2E4 N
Hence work done by that will be = 2.2E4*5 = 110 kJ
Hence the work required by the men will be the difference of the two works if the block has to be moved with a constant speed.
Wmen = 123-110 = 13 kJ
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As the work done is 13 kJ for 5 meters, the net force will be
F = 13kJ/5 = 2.6kN