Calculate the heat during the canon ball rise

[epsbanga12]

HI!

This problem has been killing me...

The Q is the following: A canon fires a canon ball which weighs 50 grams vertically at an initial speed of 600 m/s. 3 Km on top of the point where it was fired, the speed is only 50 m/s. calculate the heat during the canon ball rise.

I don't know how to use the kinetic formula in function with the energy formula, any help would be appreciated!

 

[Mattara]

The canon ball has a certain amount of kinetic energy from the start and a certain kinetic energy when the speed is 50 m/s (which is less). Some of the kinetic energy has been transformed to potential energy. The rest of the energy has been "lost" as heat etc.

Start: W_k
Finish W_k + W_p + W_h

Lets call the heat energy W_h in this case although it isn't exactly correct index.

 

[epsbanga12]

Thanks for clearing things up, Mattara.

So what would we eventually have to do to find Q, the heat produced?

 

[Hootenanny]

Using Mattara's notation Q = W_h.

~H

 

[epsbanga12]

so it would be Wh + Wp = -Wh ?

 

[Hootenanny]

Not quite, intially you have some kinetic energy. At the 'end' you have some kinetic energy, some potential and the rest as heat, therefore;

Initial Kinetic = Final Kinetic + Potential Energy + Heat

\frac{1}{2}mv_{i}^{2} = \frac{1}{2}mv_{f}^{2} + mgh + Q

Can you go from here?

~H

 

[epsbanga12]

hehe thanks Hootenanny
Actually, i meant Wk + Wp= -Wh, the former Wh was a typo...

But everything is clear now, thanks a lot to both of you!