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
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?
5 answers
Bob Pursley already did this for you.
What would be the final answer and what would be the value of Q
5.00 micro C (5.00*10-6)
is Q, the charge. It is GIVEN.
No, you do it.
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.
2 answers