1111ProblemSet08_solns.pdf

1111ProblemSet08_solns.pdf - Physics 1111 Fall 2017 PS#8...

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Physics 1111 Fall 2017 PS #8 Solutions Page 1 of 4 Physics 1111 Problem Set #8 Solutions (1) Energy Lost Around Track: (a) Friction is the only force that is not acting perpendicularly to the motion of the particle; direction matters when calculating work. Since only friction is doing any work in this case, the Work-Kinetic Energy Theorem says Δ K = W tot = Δ K = W fr . We have enough information to compute the initial and final kinetic energies from the definition: K i = 1 2 mv 2 i = 1 2 (0 . 494 kg)(7 . 69 m/s) 2 = 14 . 607 J K f = 1 2 mv 2 f = 1 2 (0 . 494 kg)(6 . 46 m/s) 2 = 10 . 308 J . Thus Δ K = 10 . 308 J - 14 . 607 J = - 4 . 299 J . In other words, 4 . 30 J are lost to friction in one revolution. (b) By the definition of work, W fr = F fr d cos θ , where θ = 180 is the angle between the friction force and the direction of motion. In this case the friction is between the particle and the floor, and the motion is purely horizontal, so F fr = μ k N = μ k mg = W fr = - μ k mgd. Since we already calculated a value for the work done by friction in part (a), we can simply solve for μ k . Here d is the circumference of the circle (the distance covered in one revolution), so - μ k mgd = Δ K = μ k = - Δ K 2 πmgr μ k = 4 . 299 J 2 π (0 . 494 kg)(9 . 81 m/s 2 )(1 . 76 m) = 0 . 0802 . (c) Since the frictional force has a constant magnitude, the amount of work done by friction is the same for every revolution. We know how much energy is lost per revolution, and
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