# Accleration of a particle: calculate work and speed

1. Oct 6, 2011

### rover_dude

1. The problem statement, all variables and given/known data

Figure 7-39 (attached) gives the acceleration of a 3.0 kg particle as an applied force moves it from rest along an x axis from x = 0 to x = 9.0 m. The scale of the figure's vertical axis is set by as = 8.0 m/s2. How much work has the force done on the particle when the particle reaches (a)x = 4.0 m, (b)x = 7.0 m, and (c)x = 9.0 m? What is the particle's speed and direction (give positive answer if the particle moves along x axis in positive direction and negative otherwise) of travel when it reaches (d)x = 4.0 m, (e)x = 7.0 m, and (f)x = 9.0 m?

2. Relevant equations
Equations I used: Area of a right triangle (1/2)(a)(b) where a and b are legs and NOT the hypotenuse. Area of a rectangle l*w (length times width) W=Kf-Ki (since Ki is 0) W=(1/2)mv^2 v=sqrt((2W)/m))

3. The attempt at a solution
Okay, this is online homework and I tried 3 different times and got the same answers, but it says they are wrong I have no idea why. So here it goes.for (a) I took the area from 0-4 of what is under the lines, so I got (1/2)(1)(8)+(3*8) to get 28. for (b) Same thing: 2(1/2)(1)(8)+(3*8)-(1/2)(1)(8)-(1)8 to get 20 (c) 2(1/2)(1)(8)+(3*8)-2(1/2)(1)(8)-(2)8 to get 8 for (d) sqrt((2*28)/3) which equals 4.32 for (e) sqrt((2*20)/3) which equals 3.65 for (f) sqrt((2*8)/3) which equals 2.30

Note: for a b and c units are Joules and d e f in in m/s

I just do not understand why these are wrong. Thanks for any help!

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2. Oct 6, 2011

### ehild

Do not forget to multiply with mass when calculating force.
The length of the second segment is 3 m, you ignored it.

ehild

3. Oct 6, 2011

### rover_dude

So... would I just multiply the whole thing by the length of the segment? so for (a) {(1/2)(1)(8)+(3*8)}*4, or what would I do?

4. Oct 6, 2011

### Staff: Mentor

So far, you are determining the pieces of area under the graph, because you hope this will be useful in answering the question.

But what are you going to do with this information (these areas) once you have them? The question doesn't specifically ask for area. How are you going to use the area under an acceleration-distance graph?

Last edited: Oct 7, 2011
5. Oct 7, 2011

### rover_dude

Is the area the force it takes to move the particle? If it is then since W=f*d I would multiply the area by the length of the segment? for (a) it woul be 28*4?

6. Oct 7, 2011

### cepheid

Staff Emeritus
Nope. The area doesn't have units of force. It has units of acceleration*distance. If you multiply by mass as a previous poster suggested, you'll have units of mass*accel*distance = force*distance = work

7. Oct 7, 2011

### rover_dude

O okay, so multiply the areas by 3? So again for (a) 28*3? that makes sense. And is the way I find speed correct?

8. Oct 7, 2011

### cepheid

Staff Emeritus
Yeah, seems like you're applying the work-energy theorem correctly.

9. Oct 7, 2011

### rover_dude

Thank you all, all I had to do was multiply the area I calculated by the mass, now that I think about it it was weird how I didn't use the mass at first to calculate work. and for the last three all I had to do was what I said in the original question sqrt((2W)/3)! Again thanks for the help!!!!