>> To calculate the power developed in this block,

you need to remember that power is defined as work divided by time.

How much work is done per unit time.

Well, we want to know the net power developed in this block.

So that's the net work.

Well that's a number that we actually already calculated.

And that's work from all the forces on this block,

as we saw a moment ago came out to be about 1,636 joules.

We know that number.

But notice that I don't know the time.

That's one of the problems that we encounter when we're calculating energy.

There is no T in those equations to really tell us how long something took.

We're going to have to find another way to solve for time.

Keep in mind this is the scenario where we

have a tension at an angle of 20 degrees above a horizontal,

where we have a frictional force resisting that pull to the right.

We have a normal force, and we have an mg.

None of those forces are changing over time.

Since the force is constant,

that means that the acceleration of this block is constant.

And it's a great opportunity for us to use kinematic equations.

So to do that, I'm going to start off with forces.

Sum of forces in the x direction on this block,

because I know it accelerates in the x direction, and

that's what I'm looking for, we're equal to mass of that block times acceleration.

So looking at my scenario, right is positive, left is negative.

So I'm going to have a t,

and now I need the horizontal component of this tension here.

That means I need to use the cosine of 20 degrees,

the adjacent side of this right triangle.

And the vertical component will come up again in the moment when we

need need [INAUDIBLE] force.

So positive T cosine of 20 minus the frictional force will equal the mass

of that block times its acceleration.

As we saw earlier, T cosine of 20 remains, but I can sub in with normal

force in the coefficient of friction, mu equals mass times acceleration.

And remember, the normal force here was not just mg, as we saw earlier.

The normal force instead will be, let's see here.

Cosine 20 is still there.

[SOUND] We're going to have u mg minus the T of the vertical component.

This is where it comes up.

Because it's not just normal force mg.

It's T sine theta.

The surface doesn't have to push back as hard,

because this tension in the string is pulling up on the block.

Will equal ma.

Lots of values here, but I'm going to sub in my numbers.

The tension was 300, cosine 20 minus the coefficient of friction 0.6.

We're talking about a 30 kilogram block, g is 10.

Minus tension of 300, sin of 20 degrees again,

because the angle between that string and the horizontal,

giving us the vertical component, equals the mass 30 times acceleration.