 #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 = (kxFf) d
(0.5) (22) (0.035)^2 = (22 x 0.0350.042) 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 = (kxFf) d
(0.5) (22) (0.035)^2 = (22 x 0.0350.042) 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