A box contains machinery that can rotate. The total mass of the box plus the machinery is m = 6 kg. A string wound around the machinery comes out through
... [Show More] a small hole in the top of the box. Initially the box sits on the ground, and the machinery inside is not rotating (left side of the figure below).Then you pull upward on the string with a force of constant magnitude F = 109 N. At an instant when you have pulled d = 0.55 m of string out of the box (indicated on the right side of the figure), the box has risen a distance of h = 0.23 m, and the machinery inside is rotating.
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POINT PARTICLE SYSTEM
Check all the forms of energy that change for the point particle system during this process:
spring potential energy
gravitational potential energy
rotational kinetic energy
translational kinetic energy
What is the y component of the displacement of the point particle system during this process?
Δy = m
What is the y component of the net force acting on the point particle system during this process?
Fnet,y = N
What is the distance through which the net force acts on the point particle system?
m
How much work is done on the point particle system during this process?
W = J
What is the speed of the box at the instant shown in the right diagram?
v = m/s
Why is it not possible to find the rotational kinetic energy of the machinery inside the box by considering only the point-particle system?
The energy principle does not apply to rotating objects
A point particle doesn't have rotational kinetic energy
The mass of the machinery is not given
EXTENDED SYSTEM
The extended system consists of the box, the machinery inside the box, and the string. Check all the forms of energy that change for the extended system during this process:
gravitational potential energy
spring potential energy
translational kinetic energy
rotational kinetic energy
What is the translational kinetic energy of the extended system, at the instant shown in the right diagram?
Ktrans = J
What is the distance through which the gravitational force acts on the extended system?
m
How much work is done on the system by the gravitational force?
Wgrav = J
What is the distance through which your hand moves?
m
How much work do you do on the extended system?
Whand = J
At the instant shown in the right diagram, what is the total kinetic energy of the extended system?
Ktotal = J
What is the rotational kinetic energy of the machinery inside the box?
Krot,f = J
A chain of metal links with total mass m = 5 kg is coiled up in a tight ball on a low-friction table. You pull on a link at one end of the chain with a constant force F = 55 N. Eventually the chain straightens out to its full length L = 1.2 m, and you keep pulling until you have pulled your end of the chain a total distance d = 3.8 m (diagram is not to scale).
09-049-chain_smaller.jpg
(a) Consider the point particle system:
What is the speed of the chain at this instant?
v = m/s
(b) Consider the extended system:
What is the change in energy of the chain?
ΔE = joules
(c) In straightening out, the links of the chain bang against each other, and their temperature rises. Assume that the process is so fast that there is insufficient time for significant thermal transfer of energy from the chain to the table, and ignore the small amount of energy radiated away as sound produced in the collisions among the links.
Calculate the increase in thermal energy of the chain.
ΔEthermal = J [Show Less]