A spring has k = 88 N/m. Use a graph to determine the work needed to stretch it from x = 3.8 cm to x = 5.8 cm.
I did it through a formula rather than a graph because I found that easier...
this is what I did:
W=1/2(88N/m)(0.058m^2 - 0.038m^2)
W=44(0.00192)
W=0.08448J
Is this right? At first I though I would do: (0.058 - 0.038)^2 which = 0.0004
and then
W=1/2(88)(0.0004)
W=0.0176 J
but I was told that's (as in 0.0176) wrong. ....Why?
Also, I don't really get how to draw the graph, I know it's REALLY hard to explain that through here but any clue is appreciated!!
Thanks in advance!! :)
I did it through a formula rather than a graph because I found that easier...
this is what I did:
W=1/2(88N/m)(0.058m^2 - 0.038m^2)
W=44(0.00192)
W=0.08448J
Is this right? At first I though I would do: (0.058 - 0.038)^2 which = 0.0004
and then
W=1/2(88)(0.0004)
W=0.0176 J
but I was told that's (as in 0.0176) wrong. ....Why?
Also, I don't really get how to draw the graph, I know it's REALLY hard to explain that through here but any clue is appreciated!!
Thanks in advance!! :)
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I agree that 0.08448 J is correct.
The reason 0.0176J is wrong is that (a² - b²) is not the same as (a - b)². It that isn't clear, try multiplying (a - b)² out
(a - b)² = a² - 2ab + b²
which is clearly (I hope) not (a² - b²).
The graph you should draw (which is worth trying to understand) is a graph of the force that must be applied to stretch the spring as a function of position. Since the spring force is
F = -k x
if x = 0 at the relaxed position, then the applied force must equal and opposite
F = k x
So the graph is a straight line passing through the origin and having slope k = 88 N/m.
Because work is related for force by
W = ∫ F ∙ dx
the work done is the area under the curve from 3.8 cm to 5.8 cm. (If you haven't had calculus, I suppose you'll just need to believe that the area under the curve represents the work.) If you sketch boundaries of the area as verticals from 3.8 cm and 5.8 cm, then your area will be a shape that can be broken into two parts, a rectangle with a triangle on top. You an use your geometry formulas to find the area of the rectangle and the triangle, and add them together to find the area. Watch your units, as your area represents work, so it won't be meters times meters, but Newtons times meters.
The reason 0.0176J is wrong is that (a² - b²) is not the same as (a - b)². It that isn't clear, try multiplying (a - b)² out
(a - b)² = a² - 2ab + b²
which is clearly (I hope) not (a² - b²).
The graph you should draw (which is worth trying to understand) is a graph of the force that must be applied to stretch the spring as a function of position. Since the spring force is
F = -k x
if x = 0 at the relaxed position, then the applied force must equal and opposite
F = k x
So the graph is a straight line passing through the origin and having slope k = 88 N/m.
Because work is related for force by
W = ∫ F ∙ dx
the work done is the area under the curve from 3.8 cm to 5.8 cm. (If you haven't had calculus, I suppose you'll just need to believe that the area under the curve represents the work.) If you sketch boundaries of the area as verticals from 3.8 cm and 5.8 cm, then your area will be a shape that can be broken into two parts, a rectangle with a triangle on top. You an use your geometry formulas to find the area of the rectangle and the triangle, and add them together to find the area. Watch your units, as your area represents work, so it won't be meters times meters, but Newtons times meters.
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You are very welcome.
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Work done = 0.5*k*(x2^2-x1^2)=(0.5*88*(0.058^2-0.038… =Answer...0.08448J should be correct if you have done the calculation correctly