Tag Archives: Integral

Work – What is it?

We started talking about work/energy today. To get started we decided Work = Force times displacement. We talked about how when force and displacement are parallel or anti-parallel things are simple to calculate. We also talked about the units of work. So if

Word Done = Force times displacement the unit equals 1 Newton per Meter but not one actually says that. Instead they say 1 Joule, which is the unit of work and energy. So remember:

1 Joule = 1 Newton per Meter

Gravity can do work too.

Fgrav = -mg

displacement = change in height = -h (since the mass falls down thanks to gravity)

Work = (-mg)(-h)=mgh

But when you lift a mass, you have to apply upward force, equal to the amount gravity is pushing down with plus just a tiny bit more. So

Work = (mg)(h) = mgh

How much work did gravity do? If you think about lifting a mass at a constant speed, and how gravity pulls a mass out of the air, it is different. Gravity is getting faster and faster. So if you think about a very small bit of the work gravity is doing, and call it dW, you now have a very small bit of work, which moves the mass over a very small bit of distance.

Read this: “The very small amount of work is equal to the mass times the current velocity times the very small change in velocity. PS In the above, dv/dt is acceleration aka the instantaneous rate of change of velocity.

So, you can add up all of these little dW’s to get the amount of work done over a larger distance, that is not so very small. Adding up all of these little dW’s is called integrating. And that is what an integral is.

Read this: The sum of all of the little works is equal to the total work done throughout the process, which is equal to the sum of all of the little changes of mass, times current velocity, times the very small change in velocity. Which is equal to 1/2 mass times velocity squared. (Mostly… the professor seemed to indicate there is actually more to it than this.)

So we looked at the problem of the work done by gravity as it pulls an eraser out of the air. And we solved for velocity based on the work done by gravity.

This solution for velocity happens to be the same as if we used one of the constant acceleration formulas. I think it only works here because in this case our Force was caused by constant acceleration.

This says “ The total work done is the final amount of work, minus any work it initially had.”

Kinetic Energy of a mass m moving with a speed v

The unit of Kinetic Energy (KE) is joules.

What happens when force displace is not parallel?

Word equals Force times displacement times cosine x.

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