What is its speed in m/s when x = 8.00m

[gijungkim]

Homework Statement

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 at x = 2.00 m, what is its speed in m/s when x = 8.00 m?
Have a picture

Homework Equations

KE = 0.5m*vf^2 - 0.5 m*vi^2
W = F * d

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?

 

[Simon Bridge]

Looks like rounding off to me.

 

[Abtinnn]

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

 

[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..

 

[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 ?!

 

[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..

 

[gijungkim]

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

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

 

[gijungkim]

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 ?!

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

 

[BvU]

Looks like rounding off to me.

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

 

[Simon Bridge]

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

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

 

[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

 

[haruspex]

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

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

 

[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.

 

[SammyS]

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.

 

[haruspex]

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.

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).