A particle with a charge of -4.3 μC and a mass of 4.4 x 10-6 kg is released from rest at point A and accelerates toward point B, arriving there with a speed of 80 m/s. The only force acting on the particle is the electric force. What is the potential difference VB - VA between A and B? If VB is greater than VA, then give the answer as a positive number. If VB is less than VA, then give the answer as a negative number.

ΔV = - 3274 V

Explanation:

For this exercise we can use conservation of energy

starting point.

         Em₀ = U = q ΔV

final point

          Em_f = K = ½ m v²

energy is conserved

          Em₀ = Em_f

          q ΔV = ½ m v²

          ΔV =

let's calculate

          ΔV =

          ΔV = 3274.4 1 V

since the charge q is negative, the potential at point B must be less than the potential at point A, so the answers

          ΔV = - 3274 V

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards

+15-35=-20

going in opposite directions