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nirvana33 [79]
3 years ago
9

A person is pulled to the right on a sled (total mass of 52.9 kg) across the surface of a frozen pond with a force of 67.7 Newto

ns so that they move with a constant velocity. What is the force of friction acting on the sled?
Physics
1 answer:
FromTheMoon [43]3 years ago
4 0

Answer:

67.7 N

Explanation:

Draw a free body diagram of the person on the sled.  There are four forces:

67.7 N  to the right,

friction force to the left,

weight force down,

and normal force up.

Constant velocity means 0 acceleration.

Sum of the forces in the x direction:

∑F = ma

67.7 N − F = 0

F = 67.7 N

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Answer:

Explanation:

Given,

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part (a)

The rope is doing the work against the gravity on the skier to uplift up to the inclined surface. Therefore the work done by the rope is equal to the work done on the skier due to the gravity

\therefore W_r\ =\ W_g\ =\ 900\ J

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part (b)

  • Initial speed of the skier = v = 1.0 m/s.

Rate of the work done by the rope is power of the rope.

Power\ =\ \dfrac{\Delta W}{\Delta t}\\\Rightarrow P\ =\ \dfrac{\Delta W}{\dfrac{d}{v}}\\\Rightarrow P\ =\ \dfrac{\Delta W\times v}{d}\\\Rightarrow P\ =\ \dfrac{900\times 1.0}{8.0}\\\Rightarrow P\ =\ 112.5\ Watt

Part (c)

  • Initial speed of the skier = v = 2.0 m/s.

Rate of the work done by the rope is power of the rope.

Power\ =\ \dfrac{\Delta W}{\Delta t}\\\Rightarrow P\ =\ \dfrac{\Delta W}{\dfrac{d}{v}}\\\Rightarrow P\ =\ \dfrac{\Delta W\times v}{d}\\\Rightarrow P\ =\ \dfrac{900\times 2.0}{8.0}\\\Rightarrow P\ =\ 225\ Watt

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Answer:

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