# Homework Help: Work, kinetic energy

1. Oct 31, 2005

work, kinetic energy, (fixed pic)

Consider the system shown in the figure. The rope and pulley have negligible mass, and the pulley is frictionless. The coefficient of kinetic friction between the 8.00-kg block and the tabletop is m_k=.25 . The blocks are released from rest.
Use energy methods to calculate the speed of the 6.00-kg block after it has descended 1.50 m.
How do i calculate work on the 8kg block.

Last edited by a moderator: Apr 21, 2017
2. Nov 1, 2005

### daniel_i_l

You know the distance that it moved, now you just need the total force (friction,gravity)

3. Nov 1, 2005

### Violet

Work of friction force equals F*S, where F is a friction force - F=k*N=k*mg, S - a shift. N - is a supporting force. Two forces: N and mg (gravity) force on upper block in vertical direction. Acceleration in this direction equals zero, therefore N=mg.
If rope is non-streched, the shift of upper block equals S=1.5m too.

4. Nov 1, 2005

### Ehsan Tarkeh

The work of friction force acting on the upper mass is equal to the amplitude of the force dot the distance that the force has acted on an object so it will be equal to (mg).(m_k).(s) which will be 8g * 0.25 * 1.5 =3g
In the other hand, friction is a kind of Non-Conservative force, and its work will be equal to the change of total energy of the system.
If you consider the initial condition to zero state you will have to calculate only the total energy of the final condition. (E2)
Remember that the velocity of two mass is equal (because the rope is not stretched)
E2=K.E2 + P.E2 =1/2(M1*V*V) + 1/2(M2*V*V) –(M2)*g*h
E2=7*V*V-6*g*1.5
Therefore -3g = 7*V*V-9g and finally V=Sqr (6g/7)=2.9 m/s

Last edited: Nov 1, 2005
5. Nov 1, 2005

### plusaf

dang...

this part always confuses me... what force does the 6kg mass exert on the rope? if the coefficient of static friction of the 8kg block is 0.25, when it's moving, it'll pull "back" with how much "force," is it? if it were 8*.25, that would be 2 to the left, and the 6 is pulling down with 6, or pulling the 8kg mass to the right with 6. that's a net 4 to the right.

energy change = change in potential energy of the 6kg weight, right? 6*1.5*[some factor?... that's the part i have problem with...]

f=m*a. total of 14kg, net f=2 whatevers. => acceleration.

hmmmm...
nah, that probably doesn't help, does it?

)
+af

Last edited by a moderator: Apr 21, 2017