How much work is required to stretch a spring by an additional 4.0 cm?

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To determine the additional work required to stretch a spring by 4.0 cm after initially stretching it by 2.0 cm, the correct approach involves calculating the total work done using the spring constant derived from the initial stretch. The spring constant (k) was calculated to be 25000 N/m, leading to the total work for a 6.0 cm stretch being 450 J. The confusion arises from the misconception that the work done can simply be added from the initial stretch; however, the work is based on the area under the force-displacement graph, which is not additive in this case. The correct calculation requires using the total stretch in the work formula, confirming that the area under the curve must be considered. Understanding the relationship between force and displacement is crucial for accurately calculating work done on springs.
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Homework Statement


If it requires 5.0 J of work to stretch a particular spring by 2.0 cm from its equilibrium length, how much more work will be required to stretch it an additioanl 4.0 cm?

Homework Equations



Force done on spring: ##F_P = kx ##
Work: ##\frac {1}{2}kx^2 ##

The Attempt at a Solution



I solved for K and got 25000.
Plugging it into the equation for the work on a spring, I get ##\frac {1}{2}(25000)(0.06)^2 ## where 0.06 is the total length stretched. Doing it that way, I get the right answer.

However, why can't I use the original information (5 J) and simply add it to the remaining 4 cm? ##5 + \frac {1}{2}(25000)(0.04)^2 ## Work is area under the force/displacement graph. Therefore, both ways should work, but that's the case.
 
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Calpalned said:

Homework Statement


If it requires 5.0 J of work to stretch a particular spring by 2.0 cm from its equilibrium length, how much more work will be required to stretch it an additioanl 4.0 cm?

Homework Equations



Force done on spring: ##F_P = kx ##
Work: ##\frac {1}{2}kx^2 ##

The Attempt at a Solution



I solved for K and got 25000.
Plugging it into the equation for the work on a spring, I get ##\frac {1}{2}(25000)(0.06)^2 ## where 0.06 is the total length stretched. Doing it that way, I get the right answer.

However, why can't I use the original information (5 J) and simply add it to the remaining 4 cm? ##5 + \frac {1}{2}(25000)(0.04)^2 ## Work is area under the force/displacement graph. Therefore, both ways should work, but that's the case.
That's because ##(0.06)^2\ne(0.02)^2+(0.04)^2\ ## .
 
Your force-displacement graph should be a triangle. Mark O, A, B, C at 0, 2, 4, 6 cm on the x axis, with D, E, F on the slope above A, B, C respectively. You want area ACFD. You won't get that by adding OAD to OBE.
 
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