Force applied by Jeff is 85 N and the force applied by Chris is 60 N. Both forces are acting in the same direction to push a rock up a hill. It means that net force is equal to the sum of these two forces.
The difference of electric potential between two points is given by the formula , where <em>d</em> is the distance between them and<em> E</em> the electric field in that region, assuming it's constant.
The electric field formula is , where <em>F </em>is the force experimented by a charge <em>q </em>placed in it.
Putting this together we have , so we need to obtain the electric force the charged ball is experimenting.
On the second drop, the ball takes more time to reach the ground, this means that the electric force is opposite to its weight <em>W</em>, giving a net force . On the first drop only <em>W</em> acts, while on the second drop is <em>N</em> that acts.
Using the equation for accelerated motion (departing from rest) , so we can get the accelerations for each drop (1 and 2) and relate them to the forces by writting:
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