Two crates, one with mass 4.00 kg and the other with mass 6.00 kg, sit on the fr

Question

Two crates, one with mass 4.00 kg and the other with mass 6.00 kg, sit on the frictionless surface of a frozen pond, connected by a light rope (Figure 1) . A woman wearing golf shoes (so she can get traction on the ice) pulls horizontally on the 6.00 kg crate with a force F that gives the crate an acceleration of 2.80m/s2 .

Part A

What is the acceleration of the 4.00 kg crate?

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Part B

Use Newton's second law to find the tension T in the rope that connects the two crates.

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Part C

Which is larger in magnitude, force T or force F?

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Part D

Use part (C) and Newton's second law to calculate the magnitude of the force F.

Two crates, one with mass 4.00 kg and the other with mass 6.00 kg, sit on the frictionless surface of a frozen pond, connected by a light rope (Figure 1) . A woman wearing golf shoes (so she can get traction on the ice) pulls horizontally on the 6.00 kg crate with a force F that gives the crate an acceleration of 2.80m/s2 .

Part A

What is the acceleration of the 4.00 kg crate?

  m/s2  

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Part B

Use Newton's second law to find the tension T in the rope that connects the two crates.

T = N

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Part C

Which is larger in magnitude, force T or force F?

Which is larger in magnitude, force  or force ? force T force F

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Part D

Use part (C) and Newton's second law to calculate the magnitude of the force F.

Explanation / Answer

1. as they are conected by a light rope and it is assumed that rope dimensions does not change, then the acceleration of 4 kg crate=acceleration of 6 kg crate=2.8 m/s^2

2.tension T is the force acting on the second crate.

now as we know from newton's second law,
force=mass*acceleration

then T=4*2.8=11.2 N

3.total force acting on the 6 kg crate is F-T.

as net force on the crate postive (as acceleration is positive)

F-T>0

hence F>T

hence F is larger.

4.
F-T=6*2.8

F=11.2+16.8

F=28 N

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