# Homework Help: Incline Issue

1. Mar 14, 2006

### Jacob87411

A m1 = 48.0 kg block and a m2 = 104.0 kg block are connected by a string as in Figure P8.36. The pulley is frictionless and of negligible mass. The coefficient of kinetic friction between the 48.0 kg block and incline is 0.250. Determine the change in the kinetic energy of the 48.0 kg block as it moves from A to B, a distance of 20.0 m.

I attached the figure.

Ok so work = change in energy.
W=FD
The forces acting on the block in the X direction are Ft (force tension in the rope) and Ffr (Frictional Force). Ft = 9.8*104kg (Maybe this is wrong?) Force Friction = cos37 * 9.8 * 48 * .25

Fx=Ft-Fr= (9.8*104)-(cos37*9.8*48*.25) = 925N
W=FD=925 * 20 = 18504. This is the change in energy, however the box changes height so there is a change in PE.

Change in PE = mgh = 48*9.8*(sin37* 20) = 5661.
So change in KE = Change in total energy - change in PE=18504 - 5661 = 12843 J.

Thanks for taking time ot help

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2. Mar 14, 2006

### Staff: Mentor

As you suspect, $F_t = m_2 g$ is wrong. You'll need to use Newton's 2nd law to solve for the tension force.

Another approach is to consider the change in total energy of the system:
$$KE_i + PE_i = KE_f + PE_f + W_{friction}$$

3. Mar 14, 2006

### Jacob87411

KEi=0...do you also take PE to initially be 0 since you take point A to be of height 0 because you don't know how high it actually is at A.

If so PEf would be mgh=(48)(9.8)(sin37*20)? If so this way is much simpler

4. Mar 14, 2006

### Staff: Mentor

Two points with that approach:

(1) Sure, take PE = 0 initially. But be sure to measure the final PE with respect to the starting point of each mass.

(2) You need to use the total energy: that means the KE and PE of both masses. (For PE: One mass goes up; one goes down.)

5. Mar 14, 2006

### Jacob87411

So initial energy = final energy + Work by Friction
The initial energy is 0

The final PE for the 48kg block is then mgh=(48)(9.8)(sin37*20)
The final PE for the 104kg block is (104)(9.8)(sin37*20) because it goes the same vertical distance as the other block?

The work of friction is (cos37)(48)(9.8)*20m

So now that we know the PE and Work of Friction we can find the total kinetic energy between the two blocks?

6. Mar 14, 2006

### Staff: Mentor

OK

OK
Careful! PE of the 104kg block is negative because it goes down. That block is not on the incline, so sin37 isn't needed: If the first block moves 20m, how far must the second block fall?

OK

Right.

7. Mar 14, 2006

### Jacob87411

Ok so the PE for the 104 block is (104)(9.8)(-20)

So:
0=PE+KE+Work

0=(104)(9.8)(-20)+(48)(9.8)(cos37)(20)+(.25*cos(37)*9.8*48*20)+KE
This gives:
KE=10,992 J
Now to find it for the 48kg block we do:
(.5*48*v^2)+(.5*104*v^2) = 10,992
Solve for velocity then just plug it into (.5*48*v^2)

That all look right, thanks a lot

8. Mar 14, 2006

### Staff: Mentor

Except for that typo (that should be sin, not cos), your equation looks OK.

I haven't checked your arithmetic, but the method is OK. Here's a timesaver: Since the blocks have the same speed, their KE is proportional to their masses. So once you've found the total KE, you can find the KE of the first mass without having to first solve for the speed, just by using a ratio:
$$KE_1 = \frac{m_1}{m_1 + m_2} KE_{total}$$