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 m 6 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 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 0.59 m of string out of the box (indicated on the right side of the figure), the box has risen a distance of h 0.18 m, and the machinery nside is rotating POINT PARTICLE SYSTEM Check all the fol of cncrgy hat change for the point particl system during this process: lational kinetic energy spring potential energy gravitational potential energy a rotational kinetic energy is the y component of the displacement of the point particle system during this proce What is the y component of the net force acting on the point particle system during this process? net,y 502 v N Wh the distal gh hich th ct force acts he point particle sy How much work is done on the point particle system during this process? W 9.04 What is the speed of the box at the instant shown in the right diagram? 74 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 The mass of the machinery is not given Apoint particle doesn't have rotational kinetic energy

Explanation / Answer

1)

Work done on the system by the gravitational force

We get the distance through the gravitational force is

d=0.18m

Wgrav=-(6x9.8x0.18)

         = -10.58 J

2)

Distance through which your hand moves =0.59+0.18

                                                                    =0.77m

3)workdone by the hand

Whand=Fxd

            =109 Nx0.77m

              =83.93 J

4)

Given The distance d=0.59m

Rotational kinetic energy   is

Krot=Fxd

      =109 Nx 0.59m

=64.31 J

5)

total kinetic energy of the system is

Ktotal=Krot+translational KE

  We get Ktrans=9.04J                  

Ktotal=64.31+9.04

=73.35 J

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