Velocity from Force position graph

In summary, A 2.40 kg particle moves along the x-axis at a speed of 4.54 m/s. At x=3.54 m, the particle's velocity is 5.48 m/s.
  • #1
shrutij
25
0

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?
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
shrutij said:

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.
 
Last edited by a moderator:
  • #3
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!
 
  • #4
shrutij said:
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.
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...

shrutij said:
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..
 
  • #5
shrutij said:

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.
 
  • #6
well i thought the instantaneous velocity was the slope at that point
 

FAQ: Velocity from Force position graph

1. What is the relationship between force and velocity on a position graph?

The slope of a position graph represents velocity, and the steeper the slope, the greater the velocity. Force, on the other hand, is a measure of the push or pull on an object that causes it to accelerate. Therefore, the greater the force, the greater the acceleration and subsequently, the greater the change in velocity.

2. How can velocity be determined from a force position graph?

The velocity at any point on a force position graph can be determined by finding the slope of the tangent line at that point. This can be done by drawing a line that just touches the curve at that point and calculating the rise over run (change in position over change in time) of that line.

3. Can the velocity be negative on a force position graph?

Yes, the velocity can be negative on a force position graph. A negative velocity indicates that the object is moving in the opposite direction of the positive direction on the graph. This could happen if the object experiences a negative force, such as friction, which causes it to decelerate or move in the opposite direction.

4. How does the shape of a force position graph affect the velocity?

The shape of a force position graph can affect the velocity by indicating changes in acceleration. A straight line on the graph indicates a constant velocity, while a curved line indicates a changing velocity. A concave up curve indicates an increasing velocity, while a concave down curve indicates a decreasing velocity.

5. How does a net force of zero affect the velocity on a position graph?

If the net force on an object is zero, there will be no acceleration and the velocity will remain constant. This would be represented by a straight line on a force position graph, indicating a constant velocity. However, if there are multiple forces acting on an object, the net force may change, causing a change in velocity as shown by the slope of the graph.

Similar threads

Replies
5
Views
1K
Replies
6
Views
3K
Replies
4
Views
3K
Replies
5
Views
987
Replies
8
Views
2K
Back
Top