# Change in Kinetic Energy/ Work

1. Oct 8, 2016

### Jrlinton

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

2. Relevant equations
ΔKE=W+FcosΘd
KE=.5mv^2
Final Vel^2=initial Velocity^2+2ax

3. The attempt at a solution
So calculate the force at x=2.6 to -5.356N, set 2.6(-5.356)=1.3Vf^2-1.3(0^2) which seemed to prove a problem as the square of the final velocity was negative, but just used that negative square in the difference of kinetic energy equation and got -5.356 J- 0= -5.356 as the change in Kinetic energy or work. I am skeptical of my methods and also a little clueless on the second part of the problem.

Many thanks

2. Oct 8, 2016

### SammyS

Staff Emeritus
Please display the problem statement in a form that can be read with reasonable effort.

Perhaps, type it out.

3. Oct 8, 2016

### Jrlinton

A 2.6 kg block is initially at rest on a horizontal frictionless surface when a horizontal force in the positive direction of an x axis is applied to the block. The force is given by https://edugen.wileyplus.com/edugen/shared/assignment/test/session.quest2529127entrance1_N10035.mml?size=14&ver=1475946606896&algorithm=1&rnd=1475946607102 [Broken], where x is in meters and the initial position of the block is x = 0. (a) What is the kinetic energy of the block as it passes through x = 2.6 m? (b) What is the maximum kinetic energy of the block between x = 0 and x = 2.6 m?

Last edited by a moderator: May 8, 2017
4. Oct 8, 2016

### SammyS

Staff Emeritus
That's better. We also have:
F(x) = (4.7 - x2)i newtons.
or
$\ \vec F (x)=(4.7-x^2)\hat i \$​

What is F at x = 0? What is F at other locations?

Last edited by a moderator: May 8, 2017
5. Oct 8, 2016

### Ray Vickson

Your formula $V_f^2 = V_i^2 + 2ax$ does not work; it assumes that $a$ is a constant, which is not true in this problem.

6. Oct 9, 2016

### Jrlinton

Okay so I integrated the force to 4.7x-x^3/3. I am unsure of what to do next

7. Oct 9, 2016

### Jrlinton

I am silly. I got 6.36 J for part 1. Now part 2 is a bit more confusing