A 0.42 kg mass is initially at rest is free to move with negligible friction alo

Question

A 0.42 kg mass is initially at rest is free to move with negligible friction along the x-axis. the figure below shows he value of an applied force as a function of time.calculate the Kinetic energy K of the mass when it reaches 8s. The answer is in J.A 0.42 kg mass is initially at rest is free to move with negligible friction along the x-axis. the figure above shows he value of an applied force as a function of time.calculate the Kinetic energy K of the mass when it reaches 8s. The answer is in J.

Explanation / Answer

Horizontally, it seems pretty straightforward -- as you point out the net area under the F(x) curve is 16 J (since the axes are meters and Newtons); that is the work done by the force. Next, we need to find the work done by friction. That is going to depend on the weight of the object PLUS the vertical component of the applied force. If that is a function of time, then deriving the work done by friction will involve an integral. If the additional vertical force is constant, then finding the normal force is easy, and the force of friction follows. If the additional vertical force is 0, then Fn = .47kg·9.8m/s² = 4.61 N and Ff = 0.033 · 4.61N = 0.152 N and Wf = 0.152N · 8m = 1.22 J Then, at position=8m, when the force is 0, the kinetic energy Ek = ½mv² = Wforce - Wfriction = 16J - 1.22J = 14.78 J and you can solve for v. NOTE: This solution is valid for applied vertical force = 0. If it is not=0, you must determine what it is. I could not do so with the information provided.

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