- #1

aqryus

- 6

- 1

- Homework Statement
- A student uses a compressed spring of force constant 22N/m to shoot a 7.5 x 10^-3 kg eraser. The magnitude of the frictional force is 4.2 x 10^-2 N. How far will the eraser slide if the spring is initially compressed 3.5 cm? Use conservation of energy.

- Relevant Equations
- W = Fd

FE = kx

EPE = 1/2kx^2

The answer is .32m. I set the elastic potential energy as equal to the work, but at first I put the force in the work equation as (F elastic - F kinetic friction) times distance and rearranged.

1/2kx^2 = (kx-Ff) d

(0.5) (22) (0.035)^2 = (22 x 0.035-0.042) d

0.013475= 0.728 d

0.013475/0.728 = d

d = .185 m which is wrong

so then i tried with just the frictional force for the work equation

(0.5) (22) (0.035)^2 = (0.042) d

0.013475 = (0.042) d

0.013475/0.042 = d

d = .32 m which is right

So I'm confused how the work done by friction (0.042 x 0.32 = 0.013475J) can stop the applied force (22 x 0.035 x 0.32 = .2464J) if it is much lower. I thought that the work of friction had to be equal to the applied force work to make an object stop. What happens to all the energy left over? Why dont I use the total work in my equation? Thank you

1/2kx^2 = (kx-Ff) d

(0.5) (22) (0.035)^2 = (22 x 0.035-0.042) d

0.013475= 0.728 d

0.013475/0.728 = d

d = .185 m which is wrong

so then i tried with just the frictional force for the work equation

(0.5) (22) (0.035)^2 = (0.042) d

0.013475 = (0.042) d

0.013475/0.042 = d

d = .32 m which is right

So I'm confused how the work done by friction (0.042 x 0.32 = 0.013475J) can stop the applied force (22 x 0.035 x 0.32 = .2464J) if it is much lower. I thought that the work of friction had to be equal to the applied force work to make an object stop. What happens to all the energy left over? Why dont I use the total work in my equation? Thank you