Velocity from Force position graph

[shrutij]

Homework Statement

A 2.40 kg particle moving along the x-axis experiences the force shown, where Fmax=6.41 N, Fmin=-6.41 N, d1=1.77 m, d2=3.54 m, d3=5.31 m, and d4=7.08 m. At x=0.00 m the particle's velocity is 4.54 m/s. What is its velocity at x=3.54 m?
http://capa.physics.mcmaster.ca/figures/kn/Graph11/kn-pic1114.png

Homework Equations

Work energy theorem

The Attempt at a Solution

W=KE
Since work is just the area under the force position graph, 1/2*b*h=1/2*m(vf^2-vi^2).
I got the area of the triangle= 3.54*6.41*0.5=11.35 J. when this was equated to KE, and isolated for vf^2= 2W/m +vi^2.
from this, i got vf=5.48 m/s, which is wrong.
Any ideas?

 

[vkash]

Homework Statement

A 2.40 kg particle moving along the x-axis experiences the force shown, where Fmax=6.41 N, Fmin=-6.41 N, d1=1.77 m, d2=3.54 m, d3=5.31 m, and d4=7.08 m. At x=0.00 m the particle's velocity is 4.54 m/s. What is its velocity at x=3.54 m?
http://capa.physics.mcmaster.ca/figures/kn/Graph11/kn-pic1114.png

Homework Equations

Work energy theorem

The Attempt at a Solution

W=KE
Since work is just the area under the force position graph, 1/2*b*h=1/2*m(vf^2-vi^2).
I got the area of the triangle= 3.54*6.41*0.5=11.35 J. when this was equated to KE, and isolated for vf^2= 2W/m +vi^2.
from this, i got vf=5.48 m/s, which is wrong.
Any ideas?

first method
First find the Force acting on the body as a function of x. then try to reach till velocity. you will get velocity as a function of displacement.(Try to do it)
Second method
from work energy theorem Net Work done on a body=change in kinetic energy of the body.
your equation(vf^2= 2W/m +vi^2)is correct but you are putting wrong values in it.
Area of shaded region from x=0 to x=3.54 is not 11.35(??). (Is it correct in magnitude?what about sign of work done).

I wan to give you a tip to learn physics. hope you would like it.

Doing question always with intuition is not good way of learning physics. always try to apply physics formula from their basic form in easy as well as tough questions. If you learn physics in this way you will never feel trouble with most kinds of question.

 

[shrutij]

I'm not sure why the area of the shaded region is wrong. I found the area of the triangle, 1/2*3.54*-6.41.
I'm assuming the work done would be negative since the force is negative.
Could you explain please?
Thanks so much!

 

[vkash]

I'm not sure why the area of the shaded region is wrong. I found the area of the triangle, 1/2*3.54*-6.41[/color].

what you have written in your first post in this thread.
this is correct. put this in work energy theorem you have written. answer will smaller than 4.54. since work is done against the motion of body...

I'm assuming the work done would be negative since the force is negative.
Could you explain please?
Thanks so much!

yes that's why area is negative. force is opposing the motion in other words that is negative.
Usually area below the x-axis is taken negative and above x-axis is positive(since all this is geometrical area). here area is under the x-axis so it is negative.

similarly you can find the velocity at any point. just take area above the x-axis as positive and below that as negative..

 

[Redbelly98]

The Attempt at a Solution

W=KE
Since work is just the area under the force position graph, 1/2*b*h=1/2*m(vf^2-vi^2).
I got the area of the triangle= 3.54*6.41*0.5=11.35 J. when this was equated to KE, and isolated for vf^2= 2W/m +vi^2.
from this, i got vf=5.48 m/s, which is wrong.
Any ideas?

The problem is that you used a positive value for the work, but it is actually negative. Other than that, your method was correct.

 

[wally1500]

well i thought the instantaneous velocity was the slope at that point