The potential at location A is 417 V. A positively charged particle is released there from rest and arrives at location B with a speed vB. The potential at location C is 799 V, and when released from rest from this spot, the particle arrives at B with twice the speed it previously had, or 2vB. Find the potential at B.

Explanation:

VA = 417 V

Vc = 799 V

Let the potential at B is V.

Mass of electron, m = 9.1 x 10^-31 kg

By the formula of energy conservation

e x ΔV = 0.5 x mv²

case I :

e x ( VB - VA) = 0.5 x m x vB²

e ( V - 417) = 0.5 mvB²     .... (1)

case II:

e x ( VB - Vc) = 0.5 x m x (2vB)²

e(V - 799) = 0.5 x 4mvB²    .... (2)

Divide equation (2) by equation (1)

V - 799 = 4V - 1668

3V = 869

V = 289.7 V

Thus, the potential at B is 289.7 V.

note: the given question is incomplete.

complete question is given below.

which is an example of kinetic energy?

a. the energy stored in ethanol  

b. a hockey puck sliding across the ice  

c. a compressed spring  

d. a ball sitting at the top of a ramp

solution: the kinetic energy is the energy posses by the virtue of the motion of the object.

in the given options, a, c and d are the example of potential energy because it is the energy posses by the virtue of the rest of the object. but option b has the kinetic energy as the object is in motion.

hence, a hockey puck sliding across the ice   is the apex example of kinetic energy.

so, the option (b) a hockey puck sliding across the ice   is the correct option.