- #1
LocknLoad
- 5
- 0
Solve Work & KE Problems with Expert Help
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Homework Help Overview
The discussion revolves around solving problems related to work and kinetic energy, specifically focusing on the application of the work-energy theorem. Participants are exploring the relationship between work done and kinetic energy in a physics context.
Discussion Character
- Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation
Approaches and Questions Raised
- Participants discuss the assumption of ignoring friction and the application of the work-energy theorem. There are attempts to derive equations for work done and kinetic energy, with some participants expressing confusion about the use of integrals and seeking simpler methods. Others suggest approximating integrals through segmenting distances and estimating average forces.
Discussion Status
The discussion is active, with participants providing various approaches and questioning the methods being used. Some guidance has been offered regarding approximating work done through graphical integration, and there is acknowledgment of different strategies to find the area under a curve. No explicit consensus has been reached, but productive ideas are being explored.
Contextual Notes
Some participants note that they have not yet learned certain mathematical concepts, such as integrals, which influences their approach to the problem. There is a mention of homework rules that require showing efforts before receiving help.
- #2
AEM
- 360
- 1
Well, according to the rules you're supposed to show your efforts to solve this. Since you didn't do that, I assume you don't have any to show. I'll get you started:
(1) We'll assume friction is being ignored. Then the work done goes into generating kinetic energy.
(2) Work done is given by
[tex]W = \int \mathbf{F} \cdot d \mathbf{l}[/tex]
(3) Since the force is not constant you need the equation for the force.
That should get you started.
(1) We'll assume friction is being ignored. Then the work done goes into generating kinetic energy.
(2) Work done is given by
[tex]W = \int \mathbf{F} \cdot d \mathbf{l}[/tex]
(3) Since the force is not constant you need the equation for the force.
That should get you started.
- #3
LocknLoad
- 5
- 0
oops i forgot that xD
I tried to solve this using a work energy theorem
W=KEfinal-KEinitial
SO because it is starting at rest the KE initial is zero
THerefore the equation would be W =.5mv^2
When solving for v, I get like 10.59, not an answer near the possible ones
I tried to solve this using a work energy theorem
W=KEfinal-KEinitial
SO because it is starting at rest the KE initial is zero
THerefore the equation would be W =.5mv^2
When solving for v, I get like 10.59, not an answer near the possible ones
- #4
LocknLoad
- 5
- 0
Also we haven't learned integrals or whatever that L is in the equation you gave me, is there an easier way to start off?
- #5
AEM
- 360
- 1
LocknLoad said:
Well, there is. You can approximate the integral in the following way: Break the horizontal distance from 0 to 150 up into convenient length segments, for starters, say 15 meters wide. That will give you 10 segments. From the graph estimate the average force over that segment (I'd pick the midpoint of the segment). Multiply that value by 15 meters. That gives you a
[tex]\Delta W[/tex] for that segment. Do this 10 times, once for each segment and add them up. That's an approximation to the total work. Set that equal to the kinetic energy and solve for v.
Three comments:
(1) The more segments you take the better the approximation.
(2) I just thought of this --you can take the midpoint of your segment and substitute it into the equation of the line to get an approximate average force for that segment
(3) What you are doing here is a simple graphical integration. Keep it in mind when you study calculus
- #6
LocknLoad
- 5
- 0
Could I just find the area under the line?
- #7
LocknLoad
- 5
- 0
Awesome, I used an intergral function on my calculator to find its total work, then set that to v. Thanks a ton!
- #8
AEM
- 360
- 1
LocknLoad said:
Well, duh. Why didn't I think of that?
Answer --I was too wrapped up in the line integral idea.
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