# Calculating work done by a spring when stretched

• vinnyzwrx
In summary, the homework statement is that at a certain point, when a spring is stretched near its elastic limit, the spring force satisfies the equation F = −α x + β x3 , where α = 12 N/m and β = 890 N/m3 . Calculate the work done by the spring when it is stretched from its equilibrium position to 0.15 m past its equilibrium.
vinnyzwrx

## Homework Statement

At a certain point, when a spring is stretched near its elastic limit, the spring force satisfies the equation

F = −α x + β x3 , where α = 12 N/m and β = 890 N/m3 .

Calculate the work done by the spring when it is stretched from its equilibrium position to 0.15 m past its equilibrium.

## The Attempt at a Solution

since this is a nonlinear curve i thought that the area under the curve would equal the work done by the spring at the .15 m

so i integrated and got

-6x2 + 222.5x4

since the spring is starting at the equilibrium position, initial position is 0 and final position is .15m so

-6(.15)2+222.5(.15)4 = -.02235

this doesn't make sense since we are pulling the spring and the work should be positive right?

Yeah, but the problem asks for the work done by the spring, not the work done by the person pulling it.

ok so maybe its negative because the spring would be working to go back to its equilibrium position. My answer is wrong though, what am i doing wrong?

am i right in my approach? i really can't think of another way of going about it...

a hint would be very nice

Last edited:
if work = 1/2kx^2 then all i need to find is the constant k given the information... i don't see how i can do this because i don't know x for the given equation, it just tells me that its near its elastic limit

Work (actually, potential energy) does not equal 1/2 kx^2 in this case. That equation applies only when F = -kx per Hookes law. This spring does not follow that law. Your original solution looks correct to me, except you answered in joules , but the problem asked for the answer in milli-joules (mJ).
-0.02235 J = ____?____ mJ ? (round it off to the nearest whole number).

-22.35. Thanks a lot, I can't believe I didn't see that.

## What is the formula for calculating work done by a spring when stretched?

The formula for calculating work done by a spring when stretched is W = 1/2kx2, where W is the work done, k is the spring constant, and x is the distance the spring is stretched.

## What is the unit of measurement for work done by a spring when stretched?

The unit of measurement for work done by a spring when stretched is joules (J).

## Can the work done by a spring be negative?

Yes, the work done by a spring can be negative if the spring is stretched in the opposite direction of its natural length. This indicates that the spring is doing work on the object, rather than the object doing work on the spring.

## What factors affect the work done by a spring when stretched?

The work done by a spring when stretched is affected by the spring constant (k), the distance the spring is stretched (x), and the direction of the force applied to the spring.

## How can I use the work done by a spring when stretched in practical applications?

The work done by a spring when stretched can be used in various practical applications, such as in the design of suspension systems, shock absorbers, and elastic potential energy storage systems. It can also be used to measure the force required to stretch a spring to a certain distance, which can be helpful in industries such as manufacturing and engineering.

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