To hoist himself into a tree, a 66.0-kg man ties one end of a nylon rope around his waist and throws the other end over a branch of the tree. He then pulls downward on the free end of the rope with a force of 366 N. Neglect any friction between the rope and the branch, and determine the man's upward acceleration.

asked by Xavier

14 years ago

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1 answer

F = ma,
a = F/m = 366 N / 66 kg = 5.5 m/s^2.

answered by Henry

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Question

To hoist himself into a tree, a 66.0-kg man ties one end of a nylon rope around his waist and throws the other end over a branch of the tree. He then pulls downward on the free end of the rope with a force of 366 N. Neglect any friction between the rope and the branch, and determine the man's upward acceleration.

1 answer

F = ma,
a = F/m = 366 N / 66 kg = 5.5 m/s^2.

Ask a New Question or answer this question.

Henry · Answer

F = ma, a = F/m = 366 N / 66 kg = 5.5 m/s^2.