# Calculate work done using integral

A chain lying on the ground is 10m long and its mass is 80kg. How much work is required to raise one end of the chain to a height of 6m? So im working in MKS system since its m and kg so the work should be in joules(J) so what i did so far is

$F(x) = 8x(9.81)$

This is because of Newtons law which states that $F = mg$ so the mass should be however high the chain currently is which i represented as x multiplied by 8 since the chains mass is 8kg/m ($\frac{80kg}{10m})$ so from there i know the work formula is

\begin{align}W &= \int_a^bF(x) dx \\&= \int_0^68x(9.81)\, dx \\&= \int_0^6 78.48x\,dx \\&=[39.24x^2]^{x=6}_{x=0}\\&= 1412.64J\end{align}

However the answer my prof gave us is $144g\ J$ which i dont understand because why is there a g when the units for work in the MKS system is J?

The answer $144g\ J$ is for convenience and exactness.

Note that $144 \times 9.81 = 1412.64\$ in correspondence with the magnitude of your answer. The units are okay, if we read between the lines. Your professor was avoiding clunky notation: $(144 kg\cdot m)g$ where $g = 9.81 m/s^2$. It's just more convenient to "strip g of its units" and state that the overall units ought to be in joules.

But there's another interesting aspect. On what planet is this chain on? $g \approx 9.81 m/s^2$ near the surface of the Earth. It varies, though on other planets. Even on Earth it's still a function of radial distance from the Earth's center of mass. Though, if we don't take $g$ as a constant, the integral becomes far more complicated since $g = g(x)$.

• Oh okay i get it now for some reason i was interpreting the g in the answer as grams instead of gravity which is where the confusion set in thanks. Commented Apr 12, 2015 at 21:28

It seems as if he used $g = 10 m/s^2$ instead of $9.81$

Let's use $g = 10 m/s^2$ and see what happens:

$$F(x) = 8x(10) = 80x$$

Thus \begin{align}W &= \int \limits_0^6 80x dx \\ &= \bigg[ 40x^2\bigg]_0 ^6 \\ &= 1440 J \\ &= 144g \ J \end{align}

which provides us with the exact result he had :). So do not fear - your process was 100$\%$ correct :). He just used $g= 10 m/s^2$ for convenience sake :).