# Work and Kinetic Energy Question

1. Jun 4, 2008

### MaZnFLiP

[SOLVED] Work and Kinetic Energy Question

The Problem/Question

The amount of work to pull an archery bow back to "Full Draw" is 12 joules. The arrow has a mass of 0.0185kg.
- How fast will the arrow leave the bow?
- How high will the arrow go if it's shot straight up?
- How far would it go if it was shot at a 35 degree angle above the horizon?
- If the arrow sticks into a 2.40kg box that is suspended from the ceiling with negligible friction, how fast will the box & arrow be moving just after the collision?

Relevant Equations

W = F$$\Delta$$x = $$\Delta$$KE
$$\Delta$$KE = KE$$_{}F$$ - KE$$_{}I$$

Known Information

Work = 12 Joules
Mass = 0.0185kg
Velocity = ?

The Attempt

W = F$$\Delta$$x = $$\Delta$$KE

(12J) = KEf - KEi

KE = 0.5mv^2

0.5(0.0185kg)v^2 - 0.5(0.0185kg)v^2

(12J) = 0.5(0.0185kg)v^2 - 0.5(0.0185kg)v^2

(12J) = v^2 - v^2?

This is where I'm getting confused because if I use the equation W = F$$\Delta$$x, it doesn't give me a displacement so I cant use that. I can also try to use $$\Delta$$KE. But that doesn't work for me either because I end up getting 12J = 0 which is not what I'm even looking for. I'm trying to find velocity but the way I've been doing it doesn't work out. Could someone please help me out on this?

2. Jun 4, 2008

### rock.freak667

well all the energy stored in the bow (12J) is converted into kinetic energy. Initially it is at rest (u=0),

$$\Delta E_k=\frac{1}{2}mv^2-\frac{1}{2}mu^2$$

3. Jun 4, 2008

### MaZnFLiP

Oh! I completely forgot about that. Okay so that now means that everything after the minus sign is canceled right? so that leaves us with $$\Delta E_k=\frac{1}{2}mv^2$$

So: $$W=\Delta KE=\frac{1}{2}mv^2$$

So that would mean that
$$12J = \frac{1}{2}(0.0185kg)v^2$$

After Solving that I now got:

$$36\frac{m}{s}^2$$

After looking at the answer sheet that is the right answer. Thanks so much for the help!

Last edited: Jun 4, 2008