Need help computing force of a defibrillator

[Quinn Pochekailo]

Homework Statement

In a TV-show, an inexperienced doctor touches the electrodes of a defibrillator and gets thrown across the room. For a moment assuming that all the energy (and only the energy) from the defibrillator goes into tossing him around, what is the maximum height above ground he could reach (i.e., if he was launched straight up)? Data: mass of doctor 93 kg, capacitance of capacitor in defibrillator 110 μF, voltage 600 V.

Homework Equations

I think I use F=mgh

The Attempt at a Solution

To calculate farads, I did (110 x 10^-6) / 600 to get 1.833 e-7. I also drew a free body diagram as the force in the y upward component must exceed that of the y downward component, which is just equal to F = 93 * 9.8.

Any help with more relevant equations would be greatly appreciated. I have been racking my brain with this problem.

 

[berkeman]

Homework Statement

In a TV-show, an inexperienced doctor touches the electrodes of a defibrillator and gets thrown across the room. For a moment assuming that all the energy (and only the energy) from the defibrillator goes into tossing him around, what is the maximum height above ground he could reach (i.e., if he was launched straight up)? Data: mass of doctor 93 kg, capacitance of capacitor in defibrillator 110 μF, voltage 600 V.

Homework Equations

I think I use F=mgh

The Attempt at a Solution

To calculate farads, I did (110 x 10^-6) / 600 to get 1.833 e-7. I also drew a free body diagram as the force in the y upward component must exceed that of the y downward component, which is just equal to F = 93 * 9.8.

Any help with more relevant equations would be greatly appreciated. I have been racking my brain with this problem.

Welcome to the PF.

The equation you wrote is not correct. I think you meant to say that the potential energy that a raised mass has is PE=mgh.

This problem is best solved by equating the initial energy in the capacitor to the final gravitational potential energy of the person. Do you know the Relevant Equation for the energy stored on a capacitor in terms of the capacitance and voltage?

 

[Quinn Pochekailo]

Based off my notes from lecture, the closest equation I can find would be Energy = Q^2/2C

 

[berkeman]

Based off my notes from lecture, the closest equation I can find would be Energy = Q^2/2C

You are given capacitance and voltage in this question, and there is an equation that gives the stored energy in terms of those variables. If it's not in your notes, use Google Images or check wikipedia -- it's a very commonly-used equation.

 

[Quinn Pochekailo]

I think I know what you are referring to, and it would be C= Q/V

 

[berkeman]

I think I know what you are referring to, and it would be C= Q/V

Nope. It only involves C and V...

 

[Quinn Pochekailo]

Energy = (1/2)CV^2?

 

[berkeman]

Energy = (1/2)CV^2?

Yes, now equate that initial stored energy on the capacitor to the final PE of the person, and calculate the maximum height that they reach.

 

[Quinn Pochekailo]

Okay so my calculations are .5(110 * 10^-6)* (600^2).

The stored energy would be 19.8 volts/charge, which I divide by (9.8*93) which would give me the height in meters.

Thank you for the help berkeman!

 

[berkeman]

Okay so my calculations are .5(110 * 10^-6)* (600^2).

The stored energy would be 19.8 volts/charge, which I divide by (9.8*93) which would give me the height in meters.

Thank you for the help berkeman!

A couple tips -- be sure to list and carry your units along with the calculations. That will help you to get familiar with checking your work, and with expressing the answers in the correct units.

Can you re-write your equations carrying units along with the numbers? And stored energy should be expressed in units of Joules...

 

[Quinn Pochekailo]

Sure.

(1/2) * (110 microfarads * 10^-6) * (600 V^2) = 19.8 J

19.8 J = mgh
19.8 J = (93 kg) (9.8 m/s2) h
h = .0217 m

Thank you for the tips and advice.