# Using the conservation of energy equation

1. Oct 20, 2007

### deNsity

Here is the problem I am given,

Two blocks are connected by a string that passes over a massless, frictionless pulley. Block A, with a mass of 2.0 kg, rests on a ramp measuring 3.0m vertically and 4.0m horizontally. Block B hangs vertically below the pulley. Use g = 10 m/s^2 for simplicity.
When the system is released from rest, block A accelerates up the slove and block B accelerates straight down. When block B has fallen through a height h=2.0m, its speed is v=6.0m/s

Using Wnc = (KE + PE) - (KEo + PEo)
I am asked first to identify which term is equal to zero.

Since there is a non conservative force (tension), Wnc does not equal 0
However, the system begins from rest, so KEo = 0

I am then asked to solve for the mass of Block B using the equation above.

with KEo = 0, I know that,
Wnc = KE + PE - PEo, However, I am unsure of how to solve for m since it appears in
all the terms. How can I isolate it?

Am I simply overthinking the problem?

Thanks!!!
-density

2. Oct 21, 2007

### Staff: Mentor

Knowing the mass of A and the height fallen and velocity given for B, one should be able to determine the mass of B.

The tension does not play a role in the KE or PE. The string simply constrains A to move with B.

http://hyperphysics.phy-astr.gsu.edu/hbase/incpl.html#c1

Note that the speed of A must equal the speed of B, but A is on a slope, so does not displace vertically as B does.

Presumably the slope is frictionless?

A more general reference - http://hyperphysics.phy-astr.gsu.edu/hbase/N2st.html

3. Oct 21, 2007

### deNsity

Yes, the slope is frictionless. I realized the error with tension last night. It should be neglected because it does postive work to Block A and negative work to Block B. I believe I am running into a problem when trying to incorporate both blocks into the conservation of energy equation. Can I use the explicit data from both in the equation at the same time? If so, how do I arrange the equation?

I was under the impression that if I use,
Wnc = (KE + PE) - (KEo + PEo)
I can only use data from either Block B, or Block A, not both together. Is this true?

Additionally, if tension is ignored, there are no non conservative forces involved, and thus Wnc = 0 correct? However the question asks me to identify the one term that equals 0 in the equation of Wnc = (KE + PE) - (KEo + PEo). Since there is no work done by a non conservative force, Wnc = 0, but the sytem starts from rest, so doesnt KEo = 0 as well?

Thanks!!
density

4. Oct 21, 2007

### Staff: Mentor

Determine the correct form of the conservation of energy and realize that,

B is going down, so its PE is decreasing while A is going up to its PE is increasing and both A and B have increases in KE. Also one already knows the mass of A.

5. Oct 21, 2007

### deNsity

I understand what is happening to the KE and PE of each block, but I believe my understanding of the energy of conservation equation is limited. I reviewed my book (Cutnell Physics), but I can't find an explenation of how to use this equation with two blocks, and this seems to be what is required.

Also, for Block B

If KEo = 0 and Wnc = 0

Wnc = (KE + PE) - (KEo + PEo)
Becomes,
0 = (KE + PE) - (0 + PEo)
--> 0 = KE +PE - PEo
--> 0 = 0.5(m)(v)^2 + 0.5(m)(g)(final height) - 0.5(m)(g)(initial height)

But m (the mass of block b) cannot be isolated

I appologize if my inablity to see something obvious is frustrating.

Thanks!
density

6. Oct 22, 2007

### rl.bhat

KE of the block B = 1/2(Mv^2) = 1/2(Mx6^2 )= 18M.
The block B is attached to A through a string. When the block B falls through 2 m block A moves upwards through 2 m along the slope.
The length of the ramp is 5 m.If you take the sine of the angle of ramp,
you get 3/5.
When the block A and B are moving, the acceleration of both the blocks must be the same and the tension in the string must be the same.
The acceleration of the block A= [T-mg(sine of angle of ramp)]/m
= [T- 2x10x3/5]/2
=(T-12)/2
The acceleration of block B = [Mg - T]/M, where M is the mass of Block B.
There fore [10M - T ]/M= [T-12]/2. solving this we get T = 32M/(M + 2)
Work done on the block B = KE of the block B
Displacement of the block B is 2m.And force on the block B is Mg-T
(10M - T)2 = 18M
10M - 32M/(M+2) = 9M or M - 32M/(M+2)= 0 or 1 = 32/(M+2)
M + 2 = 32 or M = 30kg

7. Oct 23, 2007

### rl.bhat

You can solve the problem using conservation of energy equation.
In this case change in the total energy of A must be equal to the change the total energy of B. In A both KE and PE increase, whereas in B KE increases and PE decreases.
When B moves 2m vertically downwards A moves 2m along the ramp. The sine of the angle of ramp is 3/5. Hence A rises through a height 6/5m, and it aquirs the same velocity as that of B, i.e. 6m/s. Hence change in the total energy of A is 2x10x6/5 + 1/2(2x6^2) and that of B is -10Mx2 + 1/2(Mx6^2). When you simplify you get 24+36 = -20M + 18M = -2M. Since total energy cannot be negative we get 60 = 2M or M = 30kg.

8. Oct 24, 2007