Kinetic energy of ball sliding down slope with electric field?

[asdf12312]

Kinetic energy of ball sliding down slope with electric field??

Homework Statement

As shown in the figure above, a positively charged ball is placed at point A and slides down the slope from rest. The area has a uniform electric field E = 26.1 N/C, pointing to the right. The mass of the ball is m = 27 kg and the charge is q = +4.6 C. When the ball reaches point B, it travels horizontally there after. The height from A to B is h = 23 m and the horizontal distance between A and B is d = 47 m. You can ignore friction and use g = 10 m/s2 for your calculation.

1. How much is the kinetic energy of the ball when it reaches at B (in the unit of J)?

2. If the electric field is changed to the opposite direction, How much is the kinetic energy of the ball when it reaches at B (in the unit of J) now?

3. What is the minimum strength of electric field E that can be used to stop the particle right at Point B?

Homework Equations

The Attempt at a Solution

probably need to use Ke = qEd ..? i tried but getting the wrong answer. for d i calculate using pythorean x^2+y^2=d^2 so 23^2+47^2=d^2 is that rite way to do?

 

[mfb]

Energy conservation is a good approach, but you have to consider both the electric and the gravitational energy of the ball.
The distances in those formulas are always along the direction of the force. You don't need the length of the ramp ("d").

I don't see any figure, but I can imagine how the setup looks like.

 

[rude man]

I would use d as you derived it. Then compute force along d times distance for the E and the g fields.

 

[asdf12312]

ok i have gravitation potential energy U=mgh=27*10*23=6210 but how do i use this with the Ke equation (qEd)?

 

[rude man]

ok i have gravitation potential energy U=mgh=27*10*23=6210 but how do i use this with the Ke equation (qEd)?

Compute the force along d due to the electric field, multiply by d, and you have a (negative) energy to subtract from the gravitational potential energy. The difference is your k.e. at B.

BTW qEd is not energy. It's force.

 

[deep838]

ok i have gravitation potential energy U=mgh=27*10*23=6210 but how do i use this with the Ke equation (qEd)?

next, try getting the potential energy from the electric field... add the two potential energies and get the velocity after that!

 

[deep838]

...you have a (negative) energy to subtract from the gravitational potential energy. The difference is your k.e. at B.

why negative energy? the charge is positive right?

 

[asdf12312]

Compute the force along d due to the electric field, multiply by d, and you have a (negative) energy to subtract from the gravitational potential energy. The difference is your k.e. at B.

BTW qEd is not energy. It's force.

Ke = q*E*d - -(m*g*d)

is d the same for both calculations?
Ke = d(qE+mg)

something like that?

 

[deep838]

Ke = q*E*d - m*g*d

is d the same for both calculations?
Ke = d(qE-mg)

something like that?

wait... it's Ke=qEd+mgh... not mgd...
yup that's all!

 

[asdf12312]

wait... it's Ke=qEd+mgh... not mgd...
yup that's all!

i see..so we simply use d and h given in the problem? no need to use pythagorean theorem then.

Ke=qEd+mgh=5643+6210= ~11 900 JSo I understand #2 as well, for #3 do we have to get Ke to equal 0? So then 6210/qd= ~29N/C electric field (negative in value)
did i do this rite?

 

[mfb]

i see..so we simply use d and h given in the problem? no need to use pythagorean theorem then.

Right.
The approach "rude man" described is possible, too, but it requires more calculations.

 

[rude man]

why negative energy? the charge is positive right?

Right. I was not privy to the illustration originally & imagined the ramp to go to the left & the E field to the right. This was obviously also buttressed by part c of the question ...

I also agree that this should be done via energy conservation since the ramp is not linear as I had imagined from the OP's initial submission.

 

[rude man]

i see..so we simply use d and h given in the problem? no need to use pythagorean theorem then.

Ke=qEd+mgh=5643+6210= ~11 900 J


So I understand #2 as well, for #3 do we have to get Ke to equal 0? So then 6210/qd= ~29N/C electric field (negative in value)
did i do this rite?

You have my vote!

 

[deep838]

yup...that's all for this problem...