A box contains machinery that can rotate. The total mass of the box plus the mac

Question

A box contains machinery that can rotate. The total mass of the box plus the machinery is 9 kg. A string wound around the machinery comes out through a small hole in the top of the box. Initially the box sits on the ground, and the machinery inside is not rotating (left diagram). Then yospring potential energyu pull upwards on the string with a force of constant magnitude 143 N. At an instant when you have pulled 0.47 m of string out of the box (indicated as d on the diagram), the box has risen a distance of 0.21 m (indicated as h on the diagram), and the machinery inside is rotating. POINT PARTICLE SYSTEM

1) Check all the forms of energy that change for the point particle system during this process:
rotational kinetic energygravitational potential energy translational kinetic energy

6) What is the speed of the box at the instant shown in the right diagram? v = m/s

7) Why is it not possible to find the rotational kinetic energy of the machinery inside the box by considering only the point-particle system?

Explanation / Answer

when you have pulled 0.47m, all work done is just converted to rotational kinetic energy of the machinery.
then rotational KE =143*0.47=67.21J

From 0.47m to 0.68m, assume all work done gets converted to potential energy and translational kinetic energy.

work done during lifting=0.21*143 =30.03J
potential energy=mgh=9*g*0.21=18.522 J (g=9.8m/s2)
Translational KE= work done during lifting- potential energy=11.508J

1) What is the total kinetic energy of the system? It is sum of rotational+translational energies
67.21+11.508=78.718J(Ans)
2) What is the rotational kinetic energy of the machinery inside the box?
67.21J(Ans)

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