Work Done by a Spring Force and block problem

[soul5]

Homework Statement

http://edugen.wiley.com/edugen/courses/crs1650/art/qb/qu/c07/pict_7_36.gif

gives spring force Fx versus position x for the spring–block arrangement

http://edugen.wiley.com/edugen/courses/crs1650/art/qb/qu/c07/pict_7_11.gif


The scale is set by Fs = 160 N. We release the block at x = 16.0 cm. How much work does the spring do on the block when the block moves from xi =+8.0 cm to x=+7.0 cm

Homework Equations

Fs = -kd

W = Fd

The Attempt at a Solution

Fs/d = k

160N/0.16m = k

1000 = k

kd = F

1000*1m = 1000N

1000N * 1m = W

W = 1000J


I don't know how to this question please help.

 

[LowlyPion]

When it is stretched to 8 cm how much PE is in the spring?

As it moves back to 7 cm how much potential energy is in the spring?

Isn't the difference then the Work that went to kinetic energy?

 

[Jack21222]

I disagree with your equations. Force isn't constant, so I don't think you can use W = Fd

Edit: Actually, I may be wrong... I didn't notice that graph you posted up top. I've never seen a spring problem set up with a graph like that before. Disregard this post.

 

[soul5]

When it is stretched to 8 cm how much PE is in the spring?

As it moves back to 7 cm how much potential energy is in the spring?

Isn't the difference then the Work that went to kinetic energy?

PE is just PE = 1/2kx^2

but how do we find k?


160N/0.16m = k

?

 

[LowlyPion]

PE is just PE = 1/2kx^2

but how do we find k?


160N/0.16m = k

?

That's what I would use.

F = kx so ... why not? The graph says it's linear.