- #1
shawn14parker
- 4
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I am sorry if I am doing anything wrong. I'm new to this site and physics in general. Please feel free to correct me if I'm making some stupid mistake in formatting.
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:
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)
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!
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!