- #1

shawn14parker

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1. The first figure gives spring force Fx versus position x for the spring–block arrangement of the second figure. The scale is set by Fs = 160 N. We release the block at x = 12.0 cm. How much work does the spring do on the block when the block moves from xi =+9.0 cm to (a)x=+6.0 cm, (b)x=-6.0 cm, (c)x=-9.0 cm, and (d)x=-10.0 cm?

2. Homework Equations

I am probably missing an equation here, but here are the ones I used:

F = kx

Work = -.5k(Δx)^2

Work = -.5k(Δx)^2

3.

Okay, first I converted everything to meters, which just ended up moving the decimal two places left.

Second, I found k. (Force is Kg*m/(s*s), x is meters, so k is in Kg/(s*s), which is kinda a weird unit, but whatever)

F = kx

k = F/x

k = 160/.02

k = 8000

k = F/x

k = 160/.02

k = 8000

Third, I plugged in the numbers for the work equation.

Work = -.5k Δx

-.5k = -.5*8000 = -4000

Δx = xf - xi

Work = -4000(xf - .09)^2

Lastly, I plugged in the four ending points. (x's are in meters)

Axf = .06-----> Wa = -3.6 Joules

Bxf = -.06----> Wb = -90 Joules

Cxf = -.09----> Wc = -129.6 Joules

Dxf = -.1-----> Wd = -144.4 Joules

Okay, as I was writing this up, I found out that I had made a stupid mistake in my calculations (I didn't square delta x in the work equations). However, I'm on my last shot on this online homework, and I want to get it RIGHT. So are these values correct? Or should they be positive? (I thought Joules were always positive, but I don't think I missed a minus anywhere...)

Thank you for your help!