A puck (mass m 1 = 2.70 kg) slides on a frictionless table as shown in the figur

Question

A puck (mass m1 = 2.70 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 = 8.1 kg. The mass m2 is initially at a height of h = 8.1 m above the floor with the puck traveling in a circle of radius r = 2.16 m with a speed of 8.1 m/s. The force of gravity then causes mass m2 to move downward a distance 0.81 m.

(a) What is the new speed of the puck?
m/s

(b) What is the change in the kinetic energy of the puck?
J

Explanation / Answer

(a) V = rw (w = omega = angular velocity)

here w in circular motion remain constant so

V / r = constant

So we can write

V1 / r1 = V2 / r2

here
V1 = 8.1 m/s
r1 = 2.16 m
r2 = 2.16 - .81 = 1.35 m

Now

V2 = V1 x r2 / r1

V2 = (8.1 x 1.35) / (2.16)

V2 = 5.0625 m/s (Answer)

(b) Del KE = Initial KE - FInal KE

del KE = 1/2 m1 v1^2 - 1/2 m1 v2^2

del KE = 1/2 m1 (v1^2 - v2^2)

del KE = .5 x 2.70 x ((8.1)^2 - (5.0625)^2)

del KE = 53.97 J (Answer)

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