# The work done by a spring over a specific distance

I cannot seem to find the answer for my question. I know that the formula for work is $W = FD$.

Given a mass that is attached to unstretched spring and is pushed by some force to stretch the spring, I need to calculate the work done by this force. The spring is fixed to a wall and attached to a mass.

The mass comes to a momentary stop after the mass moves $30\:\mathrm{cm}$.

So W (total) = change in kinetic energy

Change in kinetic energy will be equal to zero, since it starts from the rest and stops after $30\:\mathrm{cm}$

The work done by the force is $F \times 0.30\:\mathrm{m}$.

And the formula for the work done by a spring is $\frac{kx^2}{2}$.

I have two questions:

1) will the $x$ in the formula of work done by a spring be equal to the distance travelled by a mass? If the mass is attached to a spring, does that mean that the distance it will travel will be equal to this "x" in the formula $\frac{k x^2}{2}$?

2) If I want to find the work done by a spring, from the general formula $W = FD$

Will the work done by a spring be equal to $\frac{k x^2}{2} * 0.30\:\mathrm{m}$ (distance it will travel)?

I cannot seem to find the answer for my question. I know that the formula for work is $W=FD$.

No, not really, not in most cases and not in this one. For a force vector $\vec{F}$ acting over an infinitesimal displacement vector $\vec{dx}$ the infinitesimal work done is $dW$:

$$dW=\vec{F}\vec{dx}.$$

In your case both vectors are on the same line and point in the same direction, so the above then simplifies to:

$$dW=Fdx.$$

For a Hookean spring:

$$F=kx,$$

with $k$ the spring constant, so:

$$dW=kxdx.$$

For a total displacement $x$:

$$W=\int_0^W dW=\int_0^x kxdx=\frac{kx^2}{2}.$$

Change in kinetic energy will be equal to zero, since it starts from the rest and stops after $30\:\mathrm{cm}$

No. Initially the mass moves, then it is stopped by the spring and loses its kinetic energy. So there is a net change (loss) in kinetic energy.

2) If I want to find the work done by a spring, from the general formula $W=FD$

No. As shown above that would only work if $F$ is constant over the displacement interval $D$. Here: $W=\frac{kx^2}{2}$ with $x=0.3\:\mathrm{m}$.

• But the object starts from rest and then some force F pushes this object and then and it stops after 30 cm, shouldn't be the kinetic energy equal to 0? – Jack Oct 28 '15 at 2:36
• @Jack: when the mass is released, yes initially it speed is zero but because it undergoes acceleration $g$ (gravity), immediately it acquires speed and kinetic energy which is then converted to potential energy until the mass comes to a halt. I'll edit this in to my answer tomorrow (it's very late here). Thank you for the upvote. – Gert Oct 28 '15 at 2:52
• Ops, sorry, didn't specify that the mass slides on a horizontal surface, so there is no incline, so as i think since gravity is perpendicular it will not contribute anything, so the change in kinetic energy will be zero – Jack Oct 28 '15 at 2:56

The formula for work done by a spring is:

$$PE_{spring} = \frac{1}{2}k(\Delta x)^2$$

If I understand the question correctly, your $\Delta x$ is the 30 centimeters, or 0.3 meters. If you want to use the elastic potential energy approach, then you have to know the spring coefficient.

1) Yes, assuming the distance traveled by the mass is the same as the distance compressed/stretched by the spring.

2) The $\frac{kx^2}{2}$ is the formula for elastic potential energy done by the spring, not force, so no. I think you are confusing force and work. You can't solve using the $W = Fd$ approach based on the information you have presented because two of the three variables are unknown. Both $Fd$ and $\frac{kx^2}{2}$ are equal to work, and neither are solvable without knowing the spring coefficient $k$.

I think it is important that you develop a good conceptual understanding of spring block system as it applies to work done.

We often deal with forces that vary as they are applied; example - force impressed by a spring on a box attached to it as the box is pulled away from the spring (try this and you’ll feel more force required to pull the box as you keep pulling away. This happens because the force in the spring is increasing with the displacement that you are causing). With such variable force we cannot find the work done because the equation $W = F \cdot d$ ($d$ is displacement) assumes that force is constant.

In such a case the technique we use is to take the force at some displacement $x$. (say spring force $F = 2x$, hence variable) and assume that if this force $F$ at $x$ displaces the mass by a very very small distance $\Delta x$ (not dx for now), the force does not quite change for this short distance. In-fact you can check for yourself by taking $x = 3\,\rm m$ and then taking $x = 3.00003\rm m$. The $F$ value does not change for small $\Delta x =0.00003\rm m$. At first, $F = 6\,\rm N$ and in second case $F = 6.00006\,\rm N$.

When limit of $\Delta x$ tends to zero , $\Delta x$ changes to ${\rm d}x$ (a distance so small that you cannot determine) and we make use of integral calculus and say that the work done is, $$W=\int_{x_0}^{x_f} F\,{\rm d}x$$ This value is also the area under the curve bound by the line F = 2x and the limits of x initial and x final

Thus work done (or change) is written as $F\,{\rm d}x$ and if you have 2 distinct $x$ values you can find the change in work done by way of integration. For a generic representation of force $F = kx$, the work done is $W = \frac12 kx^2$

2. You can use $W = Fd$ only of $F$ is a constant, but you see the force changes as the spring stretches or compresses so you need to use the equation $W = \frac12 kx^2$