# Homework Help: What is its speed in m/s when x = 8.00m

1. Mar 23, 2015

### gijungkim

1. The problem statement, all variables and given/known data
The graph below shows how the force on a 0.500 kg particle varies with position. If the particle has speed https://www.physicsforums.com/x-apple-ql-id://92193C91-FD4D-4A2B-AE3A-A915237F9B1D/x-apple-ql-magic/4C0641E9-916A-4CAB-8E8F-A20E6CE7E040.pdf [Broken] at x = 2.00 m, what is its speed in m/s when x = 8.00 m?
Have a picture
2. Relevant equations
KE = 0.5m*vf^2 - 0.5 m*vi^2
W = F * d

3. The attempt at a solution
So what I basically did was to get an area of the graph because it will be F * d. After getting an area (work) I assumed that I could use KE equation like
KE = 0.5m*vf^2 - 0.5 m*vi^2 = 51J
Then I got Vf as 14.5m/s. However, the answer is 15.0 m/s. Can anyone help me what I did wrong here???

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2. Mar 23, 2015

### Simon Bridge

Looks like rounding off to me.

3. Mar 23, 2015

### Abtinnn

Are the answers reliable? I got somewhere around 14.5 as well

4. Mar 23, 2015

### T Damen

You get the answer 15.0 if you take the work exerted on the particle until x=8 to be 55 (so including the x=0 till x=2 part) . But I think your calculation is actually the correct one and the answer given is wrong..

5. Mar 23, 2015

### BvU

Get the same result as you do. The 15 m/s comes out when you do $\Delta{\rm KE} = 55 = {\tfrac 1 2} mv_f^2 - {\tfrac 1 2} m\;2.23^2, \$ and I really see no reason to use 55 instead of 51 J.

Especially since the particle can't have been released (with a small nudge) at x = 0, since the same calculation then gives 4 m/s at x = 2.

I thought about starting off at some point with a negative speed, but the calculation you did is valid, no matter what happens before, so I'm puzzled ?!

6. Mar 23, 2015

### T Damen

Yeah, the work exerted on the particle before the velocity measurement cannot have contributed to the increase of velocity after the measurement. So the answer given is just wrong..

7. Mar 23, 2015

### gijungkim

I just found these questions from the internet so it's not 100 percent reliable haha

8. Mar 23, 2015

### gijungkim

Haha I think the answer is just wrong! Thank you for helping me out though! :)

9. Mar 24, 2015

### BvU

So 14.8 is wrong and 15 is correct ? We all end up at 14.5 !

10. Mar 24, 2015

### Simon Bridge

If you insist. But all things considered, it looks like the "correct" answer is in fact incorrect.
Appearances can be deceiving.

11. Jun 24, 2015

### ChrisGeophysics

Actually the answer given is correct when rounding 14.999 to three sig fig. Work is area under curve of force vs position graph. Setting 55 = 1/2mVf^2 - 1/2mVi^2. When working with five sig figs for calculations (this comes from squaring the velocity) your answer will be 14.999 m/s => 15.0 m/s

12. Jun 24, 2015

### haruspex

Read the question carefully. It says from x=2, not x=0. The work done is only 51J.

13. Jun 24, 2015

### ChrisGeophysics

Haruspex, you are correct. I made a poor assumption that the graph continued to (0,0) for I had just seen and assigned this question. Thank you. The correct work is 51J as you stated, which results in a final velocity of 14.456 m/s ==> 14.5 m/s.

14. Jun 25, 2015

### SammyS

Staff Emeritus
This this the given graph:

At a position of about x = 1.660 m, the particle has a kinetic energy of zero. It will never be to to the left of that under the influence of this as the net force. To do so would imply negative kinetic energy, and thus an imaginary speed.

15. Jun 25, 2015

### haruspex

That helps narrow down the error in the question. Most likely it was supposed to say the given speed is at x=0. The only other I can think of is that the initial speed should have been given as 2.83, making the answer (c).