Work By Springs Video Lessons

Concept

# Problem: A child applies a force F parallel to the x-axis to a 8.00-kg sled moving on the frozen surface of a small pond. As the child controls the speed of the sled, the x-component of the force she applies varies with the x-coordinate of the sled as shown in figure . Suppose the sled is initially at rest at x=0. You can ignore friction between the sled and the surface of the pond.You may want to review (Pages 183 - 189).For related problem solving tips and strategies, you may want to view a Video Tutor Solution of Motion on a curved path.(a) Use the work-energy theorem to find the speed of the sled at 5.0 m.(b) Use the work-energy theorem to find the speed of the sled at 11.0 m.

###### FREE Expert Solution

In this problem, we are given a force-distance curve. We are also required to apply the work-energy theorem. In a force-distance graph, we know that work done by the force is equal to the area under the graph.

Work-energy theorem:

$\overline{){{\mathbf{W}}}_{\mathbf{n}\mathbf{e}\mathbf{t}}{\mathbf{=}}{\mathbf{∆}}{\mathbf{K}}{\mathbf{E}}}$

(a)

We'll determine the force at 5m. The graph does not show the position of 5. We'll subdivide the length between 4 and 8 to estimate the location of 5m. We'll also estimate the amount of force, then work out the area under the curve.

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###### Problem Details
A child applies a force F parallel to the x-axis to a 8.00-kg sled moving on the frozen surface of a small pond. As the child controls the speed of the sled, the x-component of the force she applies varies with the x-coordinate of the sled as shown in figure . Suppose the sled is initially at rest at x=0. You can ignore friction between the sled and the surface of the pond.

You may want to review (Pages 183 - 189).

For related problem solving tips and strategies, you may want to view a Video Tutor Solution of Motion on a curved path.

(a) Use the work-energy theorem to find the speed of the sled at 5.0 m.

(b) Use the work-energy theorem to find the speed of the sled at 11.0 m.

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