the puck in the figure below has a mass of 0.120 kg. its original distance from
ID: 1283353 • Letter: T
Question
the puck in the figure below has a mass of 0.120 kg. its original distance from the center of rotation is 44.0 cm, and the puck is moving with a speed of 78.2 cm/s. the string is pulled downward 17.7 cm through the hole in the frictionless table. determine the work done on the puck.
the puck in the figure below has a mass of 0.120 kg. its original distance from the center of rotation is 44.0 cm, and the puck is moving with a speed of 78.2 cm/s. the string is pulled downward 17.7 cm through the hole in the frictionless table. determine the work done on the puck.Explanation / Answer
in order to find the work done we'll need to find the difference in kinetic energies
W = KEf - KEi
The initial kinetic energy we can do right away.
KE=(1/2)mv^2=(1/2)(.120 kg)(.782 m/s)2 =.037 J
Right now the puck is drawn closer to the hole in the table by 17.7 cm. So it goes from 44cm away to 26.3 cm away. This means it'll have a different angular momentum, luckily all that is conversed due to a lack of friction going on.
Angular momentum=rmvSinx
But because the radius and the direction of motion are right angles to one another we can say that angular momentum L=rmv so now we set it up such that
L(initial)=L(final) due to conservation
rmv(initial)=rmv(final) since the masses didn't change we may cancel them out
rv(initial)=rv(final) We know the initial radius and velocity, and the final radius..just need to find the final velocity so we will solve for it.
(Ri*Vi)/R(final)=V(final)
V(Final)=(.44 m*.782 m/s)/.263 m=1.31 m/s which is faster than our original velocity, something we expected.
Now that we know the new velocity we can calculate it's kinetic energy.
KEf = (1/2)mv^2 = (1/2)(.120 kg)(1.31 m/s)2 =.103 J
So therefore W = KEf - KEi = .103J - .037 J = .066 J
This is the final workdone
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