A puck (mass m1 = 0.75 kg) slides on a frictionless table as shown in the figure

Question

A puck (mass m1 = 0.75 kg) slides on a frictionless table as shown in the figure below. The puck is tied to a string that runs through a hole in the table and is attached to a mass m2 = 2.3 kg. The mass m2 is initially at a height of h = 2.3 m above the floor with the puck traveling in a circle of radius r = 0.60 m with a speed of 2.3 m/s. The force of gravity then causes mass m2 to move downward a distance 0.23 m (a) What is the new speed of the puck? m/s (b) What is the change in the kinetic energy of the puck? Need Help? Readt Talk to a Tutor

Explanation / Answer

a) first calculate the angular momentum of the puck:
L = r * m * v
r = 0.60
m = 0.75
v = 2.3
L = 1.035 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 = 0.60 - 0.23= 0.37
m = 0.75
L = 1.035
=> v = 3.73 m/s

b) the energy before is:
E = 1/2 * m * v^2
m = 0.75
v = 2.3
E = 1.98 J
energy after:
m = 0.75
v = 3.73
E =5.22 J
Thus, the change in kinetic energy is:
5.22 - 1.98 = 3.24 J

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