Question Regarding Changes in Kinetic Energy

[Chan M]

Homework Statement

1. [/B]The only force acting on a 4.0 kg body as it moves along the x-axis varies as shown in the figure. The velocity of the body at x = 0 is 7.0 m/s.
c) What is the maximum kinetic energy attained by the body between x = 0 and x = 8 m?

Homework Equations

Work = delta K
K = kinetic energy
K = 0.5 * m * v^2
v = velocity
m = mass
Work is integral of the function of Force in regards to position

The Attempt at a Solution

So in looking at the graph, 0-1 meters, K has a positive change of 2.5 J. Between 1-2 meters, K has a negative change of 2.5 J. So they cancel out, v at 0 meters equals v at 2 meters, which is 7 m/s

Also, initial kinetic energy is 98 J
98 J = 0.5 * 4kg * (7 m/s)^2
Thus, we can find the kinetic energy at 8 meters by subtracting 2 meters and starting at the 2 meter mark since K at 2 meters equals K at 0 meters because velocities are the same at these points.

And because F is constant between 2 and 8 meters, we can simply use the W = Fd equation.
So that...
F * dr = K - Ko
-5N * 6m = K - 98J
K = 68 J

I know this isn't the correct answer, I don't know why and I need to know how to get to the correct answer.

Thanks!

 

[conscience]

Hello Chan ,

You are required to find maximum kinetic energy .Does this happen at x = 8m ?

 

[Chan M]

Hello Chan ,

You are required to find maximum kinetic energy .Does this happen at x = 8m ?

Well the energy starts out at 98J, goes up then goes back down and returns to 98J at 2 meters. The kinetic energy then decreases as the velocity decreases due to the negative force. The maximum Kinetic energy is said to be over 100J but I can't see how this is possible in any way.

 

[Chan M]

I just solved the problem, thanks.