Work/Kinetic Energy: Questions 1-4 Answered

  • Thread starter Thread starter Charanjit
  • Start date Start date
  • Tags Tags
    Energy
AI Thread Summary
The discussion revolves around calculating work done on a ball swung in a vertical circle. For the first three questions, the total work done by both tension and gravity during one complete circle is determined to be 0J, as the ball returns to its starting point. The fourth question, which involves motion along a semicircle, is more complex, leading to confusion about calculating potential energy differences. Participants clarify that the work done by gravity can be calculated using the formula W = -mgΔh, where Δh is the height difference between the highest and lowest points. Ultimately, the correct answer for the work done by gravity in this scenario is found to be -25.4J.
Charanjit
Messages
48
Reaction score
0
1. Homework Statement

A ball of mass 0.805 kg is tied to the end of a string of length 1.61 m and swung in a vertical circle.

Question 1: During one complete circle, starting anywhere, calculate the total work done on the ball by the tension in the string. Ans: 0J

Question 2: During one complete circle, starting anywhere, calculate the total work done on the ball by gravity. Ans: 0J

Question 3: Repeat part (a) for motion along the semicircle from the lowest to the highest point on the path. Note: Problem 1 is part a. Ans: 0J

Question 4: Repeat part (b) for motion along the semicircle from the lowest to the highest point on the path. Ans: unknown




2. Homework Equations

W= Fnet * d * cosx




3. The Attempt at a Solution

Well the first three questions were kind of easy, since in one complete circle, you get back to the same point, so there is no work done really. The last one is a bit hard. I thought it was 0 also, but it isn't. My free body diagram is a verticle line, at the bottom is Force of gravity, and at the top is Force of tension. Can someone give me a hint on the 4th question?
 
Physics news on Phys.org
If the circle has a radius of r, then the difference between the highest point on the path and the lowest point on the path is 2r.

What is the potential difference between those two points as far as gravity is concerned?
 
So you are saying multiply the work by 2 because of semi circle?

Because the only forces are Tension and gravity, the force, F = mg = 0.805(9.8)=7.889
So W=F*D = 7.889(1.61)=12.7J

Like that? So multiply that by 2?
 
The ball, at the highest point on the circle, will have a certain potential energy. Call it U_{high}

The ball, at the lowest point on the circle, will have a certain different potential energy. Call it U_{low}

The difference between the two values is equal to the amount of work done by gravity. The path doesn't matter.
 
So how do I actually figure that out?
 
What is the potential energy at U_{high}?
What is the potential energy at U_{low}?

If you don't know, then do you know what the potential energy of an object of mass m is a distance h from the floor?
 
U=mgh
 
I think we didn't touch on this topic yet, that's why I am having a hard time. We just covered Work/Kinetic Energy. Anyways, I don't think its difficult is it?

Ok, so Uh = -7.889
Ulow would be 0 correct, because its kinetic at the bottom, and pntential at the top.
 
Did you cover that the work (W) done by a force \left(\textbf{F}\right) is given as such?

W=\textbf{F}\Delta\textbf{x}

Gravity is the force here, given by:

\textbf{F}=-mg\hat{\textbf{y}}

So in this case, \textbf{x}=h\hat{\textbf{y}}

So substitute those into the work equation:

W=-mg\hat{\textbf{y}}\Delta h\hat{\textbf{y}}
=-mg\Delta h

Now what is \Delta h?
 
  • #10
Hmm... I worked it out, and got an answer of -25.4J, and its correct. According to what I know, I just doubled my answer. And since its moving up, its negative.
 
Back
Top