Help!! A puck (mass mi = 1.90 kg) slides on a frictionless table as shown in the

Question

Help!!

A puck (mass mi = 1.90 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 m^ = 5.7 kg. The mass m2 is initially at a height of h = 5.7 m above the floor with the puck traveling in a circle of radius r = 1.52 m with a speed of 5.7 m/s. The force of gravity then causes mass rr)2 to move downward a distance 0.57 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

given data
m1=1.90 kg
m2 = 5.7 kg
h=5.7m
r=1.52m
v=5.7m/s
distance = 0.57m

we need to find (a) What is the new speed of the puck?

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

a) first calculate the angular momentum of the puck:
L = r * m * v
r = 1.52
m = 1.90
v = 5.7
L = 1.90*5.7*1.52=16.4616Js

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 = 1.52-0.57=0.95
m = 1.90 kg
L = 16.4616

16.4616=0.95*1.90*v
=> v = 9.12 m/s

b) the energy before is:
E = 1/2 * m * v^2
m = 1.90 kg
v = 5.7 m/s
E =0.5*1.90*5.7*5.7 =30.8655J

energy after:
m = 1.90 kg
v =9.12 m/s
E =0.5*1.90*9.12*9.12 =79.015J
Thus, the change in kinetic energy is:
79.015 - 30.8655 = 48.15 J

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