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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 = 190 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 (a)x=+6.0 cm, (b)x=6.0 cm, (c)x=8.0 cm, and (d)x=10.0 cm?
 one year ago
 one year ago
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 = 190 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 (a)x=+6.0 cm, (b)x=6.0 cm, (c)x=8.0 cm, and (d)x=10.0 cm?
 one year ago
 one year ago

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azolotorBest ResponseYou've already chosen the best response.1
Hooke's Law is \[F = k \delta x\] where k is the spring constant and delta x is the displacement of the spring. Work is defined as \[W =Fdcos(\theta)\] where theta is the angle between the two vectors when they are drawn tail to tail. So, first we calculate the force generated by the displacement of the spring and then from there since the force is parallel to the distance the block is traveling we get \[W=Fd\] so once the force is found you can calculate the work at each displacement.
 one year ago

dainel40Best ResponseYou've already chosen the best response.0
yes but I'm stuck I can calculate the force but I dont know what to do after. Work is force*distance
 one year ago

azolotorBest ResponseYou've already chosen the best response.1
They give you the distances so if you have the force you can calculate the work just plug into the above formula
 one year ago

dainel40Best ResponseYou've already chosen the best response.0
but wouldnt I have to subtract from my original distance x=16?
 one year ago

azolotorBest ResponseYou've already chosen the best response.1
If I am reading this correctly no because they are asking for the distance from x(i). If I misread this then yes you just need to determine the actual displacement that's all.
 one year ago

dainel40Best ResponseYou've already chosen the best response.0
yea that didn't work just by pluging it in
 one year ago

dainel40Best ResponseYou've already chosen the best response.0
and I tried subtracting the distance and I got it wrong too
 one year ago

azolotorBest ResponseYou've already chosen the best response.1
First of all make sure you are putting the right sign on everything. Also you can extrapolate from the graph since it is a line you can easily find the equation for it and find the values at the respective x values to calculate the work. Also, make sure you convert to SI units
 one year ago

dainel40Best ResponseYou've already chosen the best response.0
Could u do step A as an examples then I will follow ur steps for the rest
 one year ago

azolotorBest ResponseYou've already chosen the best response.1
The graph has a slope of 1 and you can pick the point 0,0 for ease since it passes through the origin so y=x. It wants it at +8cm so that is y=8 so 8 newtons in the negative direction of force is exerted. It went a distance of 0.08 m so 0.08m *8N =64 Joules
 one year ago
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