Kinetic Energy Methods

1. Apr 30, 2007

marckc22

1. The problem statement, all variables and given/known data

Consider the system shown. The coefficient of kinetic friction between the 8.00 kg block and the tabletop is ųk = 0.25. Neglect the mass of the rope and of the pulley, and assume that the pulley is frictionless. Use energy methods to calculate the speed of the 6.00 kg block after it has descended 2.00 m., starting from rest.

http://img154.imageshack.us/img154/3854/image1nr8.jpg [Broken]

2. Relevant equations

i have no idea. help me.

3. The attempt at a solution

i have no idea. help me.

Last edited by a moderator: May 2, 2017
2. Apr 30, 2007

marckc22

3. Apr 30, 2007

denverdoc

welcome to PF. Do you have any ideas? Hint: the kinetic energy of the two blocks comes from the potential energy of the falling block.
So, if there were no friction:
mgh=1/2(m+M)v^2 where m=6Kg, and M=8Kg but there is friction, do you know how to compute work done by friction?

Normally we ask for some thoughts or at least posting some eqns on a problem, before anyone will help.

4. May 1, 2007

marckc22

this is what i did:
first i solved for the potential energy
PE = mgh = 6(9.8)(2) = 117.6 J

then i solved for the force done on the block
F = ų*mg = (0.25)(6)(9.8) = 14.715 N

then i solved for the work done on the block
W = (14.715)(2) = 29.43

then i solved for the kinetic energy
KE = PE - W = 117.6 - 29.43 = 88.17 J

and lastly i solved for the speed
V = (88.17/0.5(6))^-2 = 5.42 m/s

am i right?

Last edited: May 1, 2007
5. May 1, 2007

gunstar-red

$$U_p=2*6*9.8=117.6 J$$
But you've used the wrong mass in the friction force
$$F_f=0.25*8*9.8=19.6N\\ W=2*F_f=39.2 J\\ E_k=U_p-W=78.4J\\$$
Your final line also contains a mistake as you have assumed that the 6kg blosk is the only object with kinetic energy, whereas both object have the same velocity.
$$E_k=\frac{1}{2}(m+M)v^2=7v^2\\ 7v^2=78.4 J\\ v=\sqrt{\frac{78.4}{7}}=3.35ms^{-1}$$

6. May 1, 2007

marckc22

thanks a lot!