Determine Work done on the puck??

[jumpingjack90]

1. The problem statement, all variables and given/known data
The puck in the figure has a mass of 0.799 kg. Its original distance from the center of rotation is 76.2 cm, and the puck is moving with a speed of 91.0 cm/s. The string is pulled downward 38.2 cm through the hole in the frictionless table. Determine the work done on the puck.


2. Relevant equations
W=fd
centripetal force=mv^2/r
KE=1/2 mv^2

3. The attempt at a solution
Not really sure how to do this problem. I tried using the formula mv^2/r and (0.799)(0.910)^2/(0.762) = 0.868 Joules, but this is not the right answer.

[Borek]

r is not 0.762 - it changes.

That's assuming I have correctly guessed what was on the image...

[jumpingjack90]

ok. I understand that the radius changes, but How exactly do you solve this problem with the equations given?

[guguma]

ok. I understand that the radius changes, but How exactly do you solve this problem with the equations given?

Initial KE - Final KE is what you are looking for. The only factor contributing to the energy change is "pulling the string". There is no potential or friction in the problem so the total change in the energy equals Initial KE - Final KE which is the work done "by the puck" stick a negative sign in front of your result and that is the work done on the puck.

[jumpingjack90]

Initial KE= (1/2)(0.799)(0.910)^2= 0.33082595 J
Final KE= VF= 0 m/s so, (1/2)(0.799)(0)^2=0
0.33082595-0=-0.33082595 J which isn't the correct answer. What am I missing in the Final KE?

[guguma]

Initial KE= (1/2)(0.799)(0.910)^2= 0.33082595 J
Final KE= VF= 0 m/s so, (1/2)(0.799)(0)^2=0
0.33082595-0=-0.33082595 J which isn't the correct answer. What am I missing in the Final KE?

The puck does not stop, you just pull the string thus reducing the radius, The force you apply by pulling the string applies no torque so angular momentum is conserved too.

mr1v1 = mr2v2

You now what r2 is

Now you can find v2 in terms of v1.

So you can find the change in kinetic energy.