Comparing work done(Please help me mark my concept)

  • Thread starter Yh Hoo
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  • #1
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For both situation, the person oh the ship is pulling the rope with a particular constant magnitude of force, F for a fixed duration, t. Because the medium of force F is a light inextensible rope, so force doing work is in the form of tension. Treat both the person and the boat as an entity and having a combined mass of m. Also the tension in the rope due to that person's pulling force is assumed to maintain for always.

Now for situation A, the object pulls the rope anchor on a fixed pole. The tension in the rope should then be equal to F. By newton's third law, the rope reaction force pulls back the object.

Situation A
# Acceleration of objects under action of force, a = F/m

# Displacement of objects for duration t, s = (1/2) (a) (t^2)
= (1/2) (F/m) (t^2)

# Work done by F, W_A= F s
= F (1/2) (F/m) (t^2)
= (1/2) (F^2/m) (t^2)

For situation B, assume both objects freely floating on the sea having equal mass, m.

Situation B
# Acceleration of objects under action of force, a = F/2m

# Displacement of object 1 for duration t, s = (1/2) (a) (t^2)
= (1/2) (F/2m) (t^2)

Total displacement occurs = 2 s
=2 ( 1/2) ( F/2m) (t^2)
= (F/2m) (t^2)

# Work done by F, W_A= F s
= F ( 1/2) ( F/m) (t^2)
= (1/2) (F^2/m) (t^2) (Same as that in situation A)


Is my calculation true?? Thanks for your verifying.
 

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Answers and Replies

  • #2
Philip Wood
Gold Member
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Yes, I think so. Each boat in (b) acquires the same KE as the single boat in (a). So twice as much work is done overall in case (b) as in (a).
 
  • #3
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Thanks a lot. U help me a lot.
 

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