On planet Tehar, the free-fall acceleration is the same as that on Earth, but there is also a strong downward electric field that is uniform close to the planet's surface. A 4.7 kg ball having a charge of 2.74 μC is thrown upward at a speed of 17.9 m/s, and it hits the ground after an interval of 3.85 s. The acceleration of gravity is 9.8 m/s^2.
What is the potential difference between the starting point and the top point of the trajectory? Answer in units of kV.
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What is the potential difference between the starting point and the top point of the trajectory? Answer in units of kV.
Will rate best rating for correct response.
Thanks!
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I think I'd start by working out the net acceleration of the ball and therefore the force due to the electric field. That will give the electric field strength, and along the way you'll get the maximum height reached, which should be enough to get your final answer.
Time to the highest point will be half of the total flight time, so 3.85/2 = 1.925 s. At that moment the (final) vertical velocity will be zero.
v = u + at
0 = 17.9 + a(1.925)
a = -9.30 m/s/s
Hmm, odd - the acceleration is *less* than that due to gravity. The only way this makes sense is if the ball has a negative charge, so that the downward electric field exerts an upward force on it. The magnitude of the acceleration due to this force is 0.5 m/s/s or 0.5 N/kg upward. (subtracting 9.3 from 9.8)
You can find the force on the ball in N using its mass: F = ma = 4.7*0.5 = 2.35 N
Now the force due to an electric field is F = qE and the charge is 2.74 X 10^-6 C, so E = F/q = 2.35/(2.74 X 10^-6) = 857,664 N/C
To find the maximum height reached, use:
v^2 = u^2 + 2ad
0^2 = 17.9^2 + 2*(-9.3)*d
d = 17.23 m
Now E = V/d, so V = dE = 17.23 x 857,664 = 14,777,551 V = 14,778 kV
Time to the highest point will be half of the total flight time, so 3.85/2 = 1.925 s. At that moment the (final) vertical velocity will be zero.
v = u + at
0 = 17.9 + a(1.925)
a = -9.30 m/s/s
Hmm, odd - the acceleration is *less* than that due to gravity. The only way this makes sense is if the ball has a negative charge, so that the downward electric field exerts an upward force on it. The magnitude of the acceleration due to this force is 0.5 m/s/s or 0.5 N/kg upward. (subtracting 9.3 from 9.8)
You can find the force on the ball in N using its mass: F = ma = 4.7*0.5 = 2.35 N
Now the force due to an electric field is F = qE and the charge is 2.74 X 10^-6 C, so E = F/q = 2.35/(2.74 X 10^-6) = 857,664 N/C
To find the maximum height reached, use:
v^2 = u^2 + 2ad
0^2 = 17.9^2 + 2*(-9.3)*d
d = 17.23 m
Now E = V/d, so V = dE = 17.23 x 857,664 = 14,777,551 V = 14,778 kV