a puck (mass m1 = 1.25 kg) slides on a frictionless table as shown in the figure
ID: 1290714 • Letter: A
Question
a puck (mass m1 = 1.25 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 = 3.8 kg. the mass m2 is initially at a height of h = 3.8 m above the floor with the puck traveling in a circle of radius r = 1.00 m with a speed of 3.8 m/s. the force of gravity then causes mass m2 to move downward a distance 0.38 m.
(a) What is the new speed of the puck?
(b) What is the change in the kinetic energy of the puck?
a puck (mass m1 = 1.25 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 = 3.8 kg. the mass m2 is initially at a height of h = 3.8 m above the floor with the puck traveling in a circle of radius r = 1.00 m with a speed of 3.8 m/s. the force of gravity then causes mass m2 to move downward a distance 0.38 m. (a) What is the new speed of the puck? (b) What is the change in the kinetic energy of the puck?Explanation / Answer
(a)
The new radius of the path of the puck is,
rnew =1.00 m - 0.38 m
= 0.62 m
From the law of conservation of angualr momentum [mvr = mvnewrnew],
the new speed of the puck is,
vnew = vr/rnew
= [3.8 m/s][1.00 m] / [0.62 m]
= 6.13 m/s
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(b)
The change in kinetic energy of the puck is,
dK = (1/2)m[vnew2 - v2]
= (1/2)(1.25 kg)[(6.13 m/s)2 - (3.8 m/s)2]
= 14.46 J
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