Problem Involving work, velocity, given Graph

[Sade]

Homework Statement

An 8kg object is moving in the positive direction of an x axis. When it passes through x=0, a constant force directed along the axis begins to act on it. The figure gives it's kinetic energy K versus position z as it moves from x=00 to x=5.0 ; K(initial)= 30 . The forces continues to act. What is v when the object moves back through x=-3m?

Homework Equations

The Attempt at a Solution

I found the slope of the line, which is the force of the object: 30J/5m= 6N

(0,30J) is the y intercept so y=mx+b would yield the linear eq f(x)=6N(x)+ 30J

I plugged x=-3m into the equation to find the work at that point: f(-3)=6N(-3m)+ 30J
W=12J

f(0)=30 J= initial work

W= .5m(vf)^2- 30J
12J= .5(8kg)(vf)^2- 30J
4J=4(vf)^2
vf= 3.2 m/s

My answer was fairly close to the solution manual's (3.5m/s) but I'm still unsure if my reasoning was correct. Thanks so much in advance

 

[TSny]

Welcome to PF!

y=mx+b would yield the linear eq f(x)=6N(x)+ 30J

Does your slope have the correct sign?
Note that the graph represents the kinetic energy as a function of x, not the work. So, when you substitute a value for x into the linear equation, you get the kinetic energy at that value of x.