Engineers are designing a system by which a falling mass imparts kinetic energy

Question

Engineers are designing a system by which a falling mass imparts kinetic energy to a rotating uniform drum to which it is attached by thin, very light wire wrapped around the rim of the drum (the figure (Figure 1) ). There is no appreciable friction in the axle of the drum, and everything starts from rest. This system is being tested on earth, but it is to be used on Mars, where the acceleration due to gravity is 3.71 m/s^2 . In the earth tests, when m is set to 16.0kg and allowed to fall through 4.00m, it gives 250.0J of kinetic energy to the drum. Part A: If the system is operated on Mars, through what distance would the 16.0-kg mass have to fall to give the same amount of kinetic energy to the drum? h=10.6m Part B: How fast would the 16.0- mass be moving on Mars just as the drum gained 250.0 of kinetic energy? I cannot figure this out! Please help.

Explanation / Answer

It's easiest to answer the second part first. Since rotational kinetic energy is the same in any gravity, the speed will be the same as it is on earth. The falling mass will have the same kinetic energy as the drum - equal and opposite reactions and all that, and Ek = 0.5 m v squared, so for Ek=200, v will be SQRT ( 200 / 0.5 * 14) = 5.345 m/s. For the first part, remember that Ek also equals force * distance, so it is the force being applied to the mass by gravity, multiplied by the distance it travels. Force is mass x acceleration. So we have Ek = m * a * d. We know that Ek and m must remain the same, so when we go to Mars and a decreases from 9.81 to 3.71, d has to increase proportionally. Therefore, the distance on Mars will be 4.50 * (9.81 / 3.71) = 11.90 m.

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