# Spring Potential Energy Problem

1. Dec 8, 2007

### meganw

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

A nonlinear spring is compressed horizontally. The spring exerts a force that obeys the equation F(x) = Ax½, where x is the distance from equilibrium that the spring is compressed and A is a constant. A physics student records data on the force exerted by the spring as it is compressed and plots the two graphs below, which include the data and the student's best fit cures?
http://img148.imageshack.us/img148/9783/graphlk8.png [Broken]

a. From one or both of the given graphs, determine A. Be sure to show your work and specify the units.
b. i. Determine an expression for the work done in compressing the spring a distance x.
ii. Explain in a few sentences how you could use one or both of the graphs to estimate a numerical answer to part (b)i for a given value of x.

(I got a and b, but I need help on c:)

c. The spring is mounted horizontally on a countertop that is 1.3 m high so that its equilibrium position is just at the edge of the countertop. The spring is compressed so that it stores 0.2 J of energy and is then used to launch a ball of mass 0.10 kg horizontally from the countertop. Neglecting friction, determine the horizontal distance d from the edge of the countertop to the point where the hall strikes the floor
2. Relevant equations
PE=.5Kx$$^{}2$$
W = $$\int$$ F dx
W=2/3 A x^3/2 <---I got these answers from the earlier questions a/b

3. The attempt at a solution

c) PE=.5Kx$$^{}2$$

.2 = .5(25)x^2
x=.1265

Conservation of energy??
Kinetic + PEgravity + PEspring = Final Kinetic + Final PEgravity + Final PEspring

Zero Kinetic Energy + (.10)(1.3)(9.8) + (.5)(25)(.1265) = .5(.10)(vf) + Zero potential energy

Last edited by a moderator: May 3, 2017
2. Dec 8, 2007

### Astronuc

Staff Emeritus
OK. The spring is oriented horizontally. At the point where ball leaves the spring, the spring's potential energy has been transformed into the ball's kinetic energy. Since the ball is traveling horizontally, there is no change in gravitational potential energy at this point, that is until the ball goes over the edge of the table. At the time the ball leaves the table, it has a horizontal velocity that is related to its kinetic energy, and it also starts into a vertical free fall.

See this reference - http://hyperphysics.phy-astr.gsu.edu/hbase/traj.html#tra11

3. Dec 8, 2007

### meganw

Thank You! =)