Accleration of a particle: calculate work and speed

[rover_dude]

Homework Statement

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?

Homework 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))

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!

 

[ehild]

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

ehild

 

[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?

 

[NascentOxygen]

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?

 

[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?

 

[cepheid]

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?

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

 

[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?

 

[cepheid]

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?

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

 

[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!