On planet Seeturn, the free fall acceleration is 9.800 N/kg but there is also a strong downward electric field that is uniform close to the planet's surface. A 2.00 kg ball having a charge of 5.00 micro C (5.00*10-6) is thrown upward at a speed of 20.1 m/s, and it hits the ground after an interval of 4.10s. What is the potential difference between the starting point and the top point of the trajectory?
What is the kinetic energy at the bottom?
Ke = (1/2) m v^2 = .5 * 2 * 20.1^2
What is the average speed going up?
Vav = 20.1 / 2 = 10.05 m/s
So how long did it take to go up?
t = 4.1/2 = 2.05 seconds upward
so how high did it go?
h = 10.05 * 2.05
so how much work did gravity do on the way up?
Wg = m g h = 2*9.8* h
so how much was used to climb the voltage potential
Ke - Wg
but we all know that is the charge times the voltage (the potential difference)
Q V = Ke -Wg
Bob Pursley already did this for you.
What would be the final answer and what would be the value of Qposted by john
5.00 micro C (5.00*10-6)
is Q, the charge. It is GIVEN.
No, you do it.
With your way, the answer is 50000 V but that is not the correct answer.posted by paul