One force represented along an x-axis

[blakester]

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

The only force acting on a 2.4 kg body as it moves along the x axis varies as shown below. The velocity of the body at x = 0 is 4.0 m/s.

(a) What is the kinetic energy of the body at x = 3.0 m?

(b) At what value of x will the body have a kinetic energy of 8.0 J?

(c) What is the maximum kinetic energy attained by the body between x = 0 and x = 5.0 m?

http://www.webassign.net/hrw/hrw7_7-41.gif

2. Relevant equations
Work is equal to the change in kinetic energy
W=F*d

3. The attempt at a solution
I'm at a loss here, I really just need some help interpreting what this graph is. I think a point in the right direction might get me started so any help is appreciated. I've been struggling a bit on work and kinetic energy and were already finishing potential energy in class so any help is appreciated heh

[blakester]

ok i'm just stabbing in the dark here but this is what i was thinking

Ke=1/2mv^2
initial Ke + W = Ke final
so if for part a x=3 and the force is -4 at that point work should be -12.

the velocity at 0 is 4 m/s so initial kinetic energy is 19.2
19.2-12=7.2 kinetic energy at x=3

[blakester]

ok i figured it out on my own and for anyone else wondering out there about this problem lemme explain.

using what i said in the before post
ke=1/2mv^2
initial Ke + W = Ke final (which i'll call Kei and Kef now)
at x=3 the work area added up under the line is 2 (at 0) 0 (at 1) -2 (at 2) and -4 (at 3). Add up those areas for a total area which gives you the work of -4. Plug that W into Kei + W = Kef, and since we already know Kei from whats givin in the problem we get the answer.

part b is working backwords using the same kei + an unknown W = 8.
what you get for work is 11.2 and with 8 W at 4 and 12 W at 5 it's just a matter finding where in between you need to be.

you should be able to get C if you got this far but if not lemme know and i'll help more