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Just took my midterm, can someone explain a problem I missed

  • Thread starter nperk7288
  • Start date
17
0
I just took my Physics Midterm and there was one problem on the test that I had no clue on and just left blank. Here's the problem:
A 6.0 kg box is attached to a rope that is wound around a pulley of radius 0.150 m and mass 10.0 kg. The box is initially at rest and is suspended 3.0 m above the ground. There is no slipping between the rope and the pulley. Given this information, find both the velocity of the box and the tension of the rope the moment before the box contacts the ground.
Anyone have any ideas?
 

OlderDan

Science Advisor
Homework Helper
3,021
1
Energy conservation will do for the velocity. Initial potential energy will be converted to kinetic energy of rotation for the pulley and translation for the box. The angular velocity of the pulley and linear velocity of the box are related by the pulley radius.

Since you need to also find the tension, you could use the velocity change to find the linear acceleration of the box and use that to find the net force on the box and deduce the tension.

Alternatively, you can write equations of motion for the box and angular motion for the pulley, both of which involve the unklnown tension. The accelerations are related by the radius. Solve simultaneous equations to find the tension and the acceleration. Use the acceleration to find the final velocity.

Give it a try.
 
17
0
Not sure if I did what you said correctly or not but here's what I've gotten:
mgh = 1/2(Iω2) + 1/2(mv2)
Then by substituting in v/r = ω I got v = 5.66 m/s
After that I found the acceleration to be 5.34 m/s2 and found the Force downward to be 32 N. Since the force without the rope would be 58.8 N I figured the tension to be 26.8 N
Is this correct?
 

OlderDan

Science Advisor
Homework Helper
3,021
1
nperk7288 said:
Not sure if I did what you said correctly or not but here's what I've gotten:
mgh = 1/2(Iω2) + 1/2(mv2)
Then by substituting in v/r = ω I got v = 5.66 m/s
After that I found the acceleration to be 5.34 m/s2 and found the Force downward to be 32 N. Since the force without the rope would be 58.8 N I figured the tension to be 26.8 N
Is this correct?
Looks good to me.
 

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