A 2.8 kg body is at rest on a frictionless horizontal air track when a horizontal force F acting in the positive direction of an x axis along the track is applied to the body. The force F is applied to the body at t = 0, and the graph records the position of the body at 0.50 s intervals. How much work is done on the body by the applied force F between t = 0 and t = 1.0 s?

Find the distance it went:

distance= 1/2 (F/m)t^2

then work is the dot product of force and distance.

asked by Technoboi11

18 years ago

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A 2.8 kg body is at rest on a frictionless horizontal air track when a horizontal force F acting in the positive direction of an x axis along the track is applied to the body. The force F is applied to the body at t = 0, and the graph records the position of the body at 0.50 s intervals. How much work is done on the body by the applied force F between t = 0 and t = 1.0 s?

Find the distance it went:

distance= 1/2 (F/m)t^2

then work is the dot product of force and distance.

Find the distance it went:

distance= 1/2 (F/m)t^2

then work is the dot product of force and distance.