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# Work and energy

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1) Evaluate the work done W= Int (from O to P) F &#61476; dr = Int (from O to P) (Fx dx + Fy dy) by the two-dimensional force F = (x2, 2x, y) along the three paths joining the origin to the point P = (1, 1) and defined as follows: (a) This path goes along the x axis to Q = (1,0) and then straight up to P. (Divide the integral into two pieces, Int (from O to P) = Int (from O to Q) + Int (from P to Q). (b) On this path y = x2, and you can replace the term dy in W by dy = 2xdx and convert the whole integral into an integral over x. (c) This path is given parametrically as x = t3, y = t2. In this case rewrite x, y, dx and dy in W and convert the integral into an integral over t.

2) A particle of mass m is moving on a frictionless horizontal table and is attached to a massless string, whose other end passes through a hole in the table, where I am holding it. Initially the particle is moving in a circle of radius ro with angular velocity wo, but I now pull the string down through the hole until a length r remains between the hole and the particle. (b) Assuming that I pull the string so slowly that we can approximate the particle's path by a circle of slowly shrinking radius, calculate the work I did pulling the string. (b) Compare your answer to part a with the particle's gain in kinetic energy.

3) (a) Consider a mass m in a uniform gravitational field g, so that the force on m is mg, where g is a show that the work done by gravity is Wgrav(1 &#61556; 2) = -mgh where h is the vertical height gained between points 1 and 2. Use this result to prove that the force of gravity is conservative (at least in a region small enough so that g can be considered constant). (b) Show that, if we choose axes with y measured vertically up, the gravitational potential energy is U = mgy (if we choose U = 0 at the origin).

4) For a system of N particles subject to a uniform gravitational field g acting vertically down, prove that the total gravitational energy is the same as if all the mass were concentrated at the CM of the system, that is: U = Summation (alpha on bottom of summation) Ualpha = MgY, where M = summation of Malpha is the total mass and R = (X, Y, Z) is the position of the CM, with the y coordinate measured vertically up.

5) Consider a small frictionless puck perched at the top of a fixed sphere of radius R. If the pluck is given a tiny nudge so that it begins to slide down, through what vertical height will it descend before it leaves the surface of the sphere? [Hint: Use conservation of energy and Newton's second law.]

6) (a) The force exerted by a one-dimensional spring, fixed at one end, is F = -kx, where x is the displacement of the other end from its equilibrium position. Assuming that this force is conservative (which it is) show that the corresponding potential energy is U = (1/2) kx2, if we choose U to be zero at the equilibrium position. (b) Suppose that this spring is vertically hung from the ceiling with a mass m suspended from the other end and constrained to move in the vertical direction only. Find the extension xo of the new equilibrium position with the suspended mass. Show that the total potential energy (spring plus gravity) has the same form (1/2) ky2 if we use coordinate y equal to the displacement measured from the new equilibrium position at x = xo (and redefine our reference point so that U = 0 at y = 0).

See attached file for full problem description.