A puck (mass m1 = 2.50 kg) slides on a frictionless table. The puck is tied to a

Question

A puck (mass m1 = 2.50 kg) slides on a frictionless table. The puck is tied to a string that runs through a hole in the table and is attached to a mass m2 = 7.5 kg. The mass m2 is initially at a height of h = 7.5 m above the floor with the puck traveling in a circle of radius r = 2.00 m with a speed of 7.5 m/s. The force of gravity then causes mass m2 to move downward a distance 0.75 m. What is the new speed of the puck? What is the change in the kinetic energy of the puck?

Explanation / Answer

a) first calculate the angular momentum of the puck: L = r * m * v r = 3.88 m = 4.85 v = 14.6 L = 274.75 Js As there is no net torque working on the system, angular momentum is conserved, thus we can calculate the new velocity, with, again: L = r * m * v r = 3.88 - 1.46 = 2.42 m = 4.85 L = 274.75 => v = 23.4 m/s b) the energy before is: E = 1/2 * m * v^2 m = 4.85 v = 14.6 E = 516.9 J energy after: m = 4.85 v = 23.4 E =1372.8 J Thus, the change in kinetic energy is: 1372.8 - 512.9 = 811 J

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