Work-Energy Theorem rock throw Question

[hollystella]

This is the problem:
You throw a 20-N rock vertically into the air from ground level. You observe that when it is 15.0 m above the ground, it is traveling at 25.0 m/s upward. Use the work-energy theorem to find a) its speed just as it left the ground; b) its maximum height.

I drew a free-body diagram with an unknown force F pointing upwards, and the mg = 20 N pointing down. I also found the mass of the object by using 20 N = mg to get m = 2.04 kg.

I know that the work theorem is W(total) = K2 - K1, with K = (1/2)m(v)^2. I figured out K2 with the given 25.0 m/s and the 2.04 kg, which came out to be 637.5 J. Because I'm trying to find the initial speed, K1 looked like this = (1/2)(2.04 kg)(v1)^2. I tried to find out what W(total) was to complete this problem, but then became stuck because, in W = Fs, I didn't know what F was. I couldn't calculate F = ma in the vertical direction because I didn't have an acceleration. How do I go about figuring out W(total) from this information?

Thanks!

 

[aaroman]

Why there should be an unknown force F pointing upwards as the rock is rising?

 

[hollystella]

I was thinking there would be a force upwards because it was thrown up into the air.

Is there another way I should be looking at this?

 

[Doc Al]

I drew a free-body diagram with an unknown force F pointing upwards, and the mg = 20 N pointing down.

As soon as the rock leaves contact with your hand, the only force on it is gravity (ignoring air resistance, of course).

 

[hollystella]

I see...

So then W(total) = (20 N) * (15 m)? Then W(total) would be 300 J. Then I would be able to put that into W(tot) = K2 - K1, find what K1 is and then equate that to (1/2)m*(v1)^2 to find the initial speed.

Is that right?

 

[Doc Al]

Yes, but realize that the work done on the rock is negative since the force (down) and the displacement (up) are in opposite directions.