Rate of Gravitational Force (Power)

[Lori]

Homework Statement

At what rate is the gravitational force doing work on a 2.0 kg projectile at an instant when the velocity of the projectile is 4.0 m/s directed 30 degrees above the horizontal.

the answer is -39 W but i need help on why!

Homework Equations

P = W/T
Power = Work*V
=Force*cos(theta)*V

The Attempt at a Solution

I know that the force of the gravity is just 9.8*2, but why is the angle that i plug in cos(120) ? I'm not sure why/how we use angle of 120 as oppose to 90-30 = 60 degrees...

If you can provide a sketch for me visually, it would help a lot...

Here's what i thought:

 

[phyzguy]

How does cos(60) compare to cos(120)? This might help answer your question.

 

[Lori]

How does cos(60) compare to cos(120)? This might help answer your question.

Ohh, it's just the negative of it... but i get why work would be negative in this case, but how can i draw it so that i don't mess up the angles next time? Since i have to do it right on the exam, i just draw the diagram and don't think of the signs! Is there a actual sketch of it that it should look like, so that i can learn how to draw it?

 

[CWatters]

Look at the definition of work done. Eg.. W=f.s

The angle between the velocity and force of gravity vectors is 120 degrees when they are drawn from the same point.

 

[Lori]

Look at the definition of work done. Eg.. W=f.s

The angle between the velocity and force of gravity vectors is 120 degrees when they are drawn from the same point.

Ohhhh. I see it now. Thanks

 

[CWatters]

This is the way I would draw it...