Find work given a pulley and two weights

In summary, the problem asks for the work done when two blocks are connected by a very light string. The first weight is 12.0 N and the second weight is 20.0 N. The problem states that the tension in the string is not included in the problem. The first weight is set to move and the second weight is set to stay still. The problem states that the work done is 5.63 J.
  • #1
fishturtle1
394
82

Homework Statement



Problem 6.65
Two blocks are connected by a very light string passing over a massless and frictionless pulley (Figure (Figure 1) ). The 20.0-N block moves 75.0 cm to the right and the 12.0-N block moves 75.0 cm downward.

Question: Find the total work done on 20.0-N block if there is no friction between the table and the 20.0-N block.

Homework Equations


ΣF = ma
Wtotal = F * s cos(φ), where the force is constant.
Wtotal = 1/2(m)vf^2 - 1/2(m)vi^2
vf^2 = vi^2 + 2ad

The Attempt at a Solution


So I started with the 12.0N weight.

ΣF = m1a = T - 12.0N, where m1 is the mass of the 12.0N weight. I also say acceleration is the same for both weights because they are in the same system.

then T = (12/9.8)a+12.0N

Then I set up my force equation for the 20.0N weight.

ΣF = m2a = T, where m2 = 20/9.8

(20/9.8)a = (12/9.8)a + 12.0N
(8/9.8)a = 12.0N
a = 14.7 m/s^2

Then I plug this into my equation for tension.

T = (12/9.8)14.7 + 12.0
T=18+12.0
T=30N

Then W = T * s cos(0°)
W = (30)(.75)(1) = 22.5 J.

Which is wrong.

2nd attempt

I thought maybe force is not constant, although I'm not sure why it wouldn't be. SO i used this equation.

W = 1/2(m)vf^2 - 1/2(m)vi^2, where i assume vi = 0m/s
then
W = 1/2(m)vf^2

to solve for vf: vf^2 = vi^2 +2ad
vf^2 = 2(14.7)(.75)
vf^2 = 22.05

Then
W = 1/2(20)(22.05)
W = 220.5J

which is also wrong. I know my displacement vector is correct so it must be my acceleration but I can't see anywhere that i made an error in calculating it.
 
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  • #2
Hi,

I take it figure 1 is like here ?

Check a
 
  • #3
The question statement is unclear. Are we to assume that the blocks are released from rest, so they will accelerate in the obvious way, but you are just to find the work done over the first 75cm of movement?
fishturtle1 said:
m1a = T - 12.0N,
Looks like you are defining positive up, so a is expected to be negative.
fishturtle1 said:
m2a = T,
But now a needs to be positive.
 
  • #4
BvU said:
Hi,

I take it figure 1 is like here ?

Check a

Yes that is the figure, sorry i forgot to add it. I'll add it now. Never mind i waited too long to edit. But yes that is the figure.
 
  • #5
haruspex said:
The question statement is unclear. Are we to assume that the blocks are released from rest, so they will accelerate in the obvious way, but you are just to find the work done over the first 75cm of movement?

Looks like you are defining positive up, so a is expected to be negative.

But now a needs to be positive.
I am unsure if we are supposed to assume the blocks start from rest. I think what that means is the initial velocity and final velocity aren't needed to solve this problem.

I also did away with my original equations for tension after looking at http://hyperphysics.phy-astr.gsu.edu/hbase/hpul2.html. This website had a formula for tension which was a = [(mass of 12.0N weight)(g) - μ(mass of 20.0N weight)(g)]/(mass of 12.0N + mass of 20.0N). I used this to solve for acceleration. Then plugged that acceleration in for :

ΣF = (mass of 20.0N)(a) = T,
(20/9.8)(3.675) = T,
T = 7.501

Then i plugged this force into my work equation:

W = F * s cos(Φ)
W = 7.501 * .75 * 1 = 5.63J. This came up as the correct answer.But I'm stuck on how we used the original equation to find acceleration.

Their vector equation, I assume, looked like this:

ΣF = (m1 + m2)a = w2 - ƒ - T2 + T1, where T1 = T2.

so my confusion is this: I was under the impression you could only use Newton's second law for one object at a time. So in this case are we just setting up a Newton's second law equation twice and then combining them?

No friction.
1st weight: ΣF = m1a = T
2nd weight: ΣF = m2a = w2 - T,
Then their sum would be: m1a + m2a = w2 - T + T => (m1 + m2)a = w2 = > a = w2/(m1+m2).

I guess this is the first time I've seen this happen so I'm not sure if this is the calculations made.
 

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  • #6
fishturtle1 said:
are we just setting up a Newton's second law equation twice and then combining them?
Yes, you can get the same equation by doing that.
As I hinted before, you had a sign error in your original equations. I cannot say exactly where the error was, just that the equations were inconsistent.
 
  • #7
haruspex said:
Yes, you can get the same equation by doing that.
As I hinted before, you had a sign error in your original equations. I cannot say exactly where the error was, just that the equations were inconsistent.
ok. Thank you for your help! I will do some more two weights/one pulley problems to make sure i got it
 

1. How do you determine the work done on a pulley system with two weights?

The work done on a pulley system with two weights can be determined by calculating the force applied and the distance moved. The work done is equal to the force applied multiplied by the distance moved in the direction of the force. This can be represented by the formula W = Fd, where W is the work done, F is the force applied, and d is the distance moved.

2. What is the relationship between the weight of the objects and the work done in a pulley system?

In a pulley system, the work done is directly proportional to the weight of the objects being lifted. This means that the heavier the weights, the more work is required to lift them. This relationship can be seen in the formula W = Fd, where W is the work done, F is the force applied, and d is the distance moved.

3. Can the work done on a pulley system be negative?

Yes, the work done on a pulley system can be negative. This occurs when the force applied and the distance moved are in opposite directions. In this case, the work done is considered to be negative, indicating that energy is being transferred out of the system.

4. How does the number of pulleys affect the work done in a pulley system?

The number of pulleys in a pulley system does not affect the work done. The work done is still determined by the force applied and the distance moved, regardless of the number of pulleys used. However, using multiple pulleys can make it easier to lift heavy weights by reducing the amount of force needed.

5. Is the work done on a pulley system affected by the direction of the force applied?

No, the work done on a pulley system is not affected by the direction of the force applied. As long as the force is being applied and the weights are being lifted, work is being done. The direction of the force only affects the direction of the movement, not the amount of work done.

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