Finding Kinetic Energy From A Graph

[danest]

Homework Statement

A 2.0 kg lunchbox is sent sliding over a frictionless surface, in the positive direction of an x-axis along the surface. Beginning at time t = 0, a steady wind pushes on the lunchbox in the negative direction of the x axis. Figure 7-50 shows the position x of the lunchbox as a function of time t as the wind pushes on the lunchbox. From the graph, estimate the kinetic energy of the lunchbox at (a)t = 1.0 s and (b)t = 5.0 s. (c) How much work does the force from the wind do on the lunchbox from t = 1.0 s to t = 5.0 s?
http://edugen.wiley.com/edugen/courses/crs1650/art/qb/qu/c07/fig07_50.gifFig. 7-50

Homework Equations

Change in Y/ Change X

K = .5mv^2

The Attempt at a Solution

I tried finding the slope for each part bc that would be the velocity and then I plug in the velocity into the K equation but it did not work. I am not sure what else to try after that.

 

[kuruman]

Your method is correct. Show us exactly what you did and how you put numbers in the equations. We can't pinpoint what didn't work without knowing what you did.

 

[danest]

For part a) I used the the coordinate (0,0) and (1,1) to get the slope and it was just 1
Then I plug it into K = (1/2)(2)(1^2) = 1

For part b ) I used the coordinates (0,0) and (5,2.5) and found the slope for that to be (1/2)
then K = (1/2)(2)(1/2^2) = .25

I think for part 3 you have to subtract the first two answers to get the total but I can't get those yet

 

[kuruman]

For part a) I used the the coordinate (0,0) and (1,1) to get the slope and it was just 1
Then I plug it into K = (1/2)(2)(1^2) = 1

That's about right.

For part b ) I used the coordinates (0,0) and (5,2.5) and found the slope for that to be (1/2)

You can't do that. The slope is a line tangent to the graph at the appropriate point. At 5.0 s, the line tangent to the graph is parallel to the time axis. What is the slope at t = 5.0 s?

 

[danest]

The slope would just be 0 then.

I also tried the answer for a and it was wrong. Is there something I missed there?

 

[kuruman]

The slope would just be 0 then.

I also tried the answer for a and it was wrong. Is there something I missed there?

Not really, but you may wish to refine the estimate for the velocity. Read the position at 0.5 s and at 1.5 s, take the difference and divide by (1.5 - 0.5) s. You should get a better value that way.