On planet Tehar, the free-fall acceleration is the same asthat on Earth, but the

Question

On planet Tehar, the free-fall acceleration is the same asthat on Earth, but there is also a strong downward electric fieldthat is uniform near the planet's surface. A 2.0 Kg ball with acharge of 5.0 C is thrown upward at a speed of 20.1 m/s. Ithits the ground after an interval of 4.1 sec. What is the potentialdifference between the starting point and the top point of thetrajectory? NOTE: The Cramster solution is not even close to the correctanswer, so only answer if you really know your Physics. Not a textproblem. I know that youn can use kinematics and V= QLD tosolve. Moderators response was incorrect. Answer must include all givenvalues or no credit will be issued. On planet Tehar, the free-fall acceleration is the same asthat on Earth, but there is also a strong downward electric fieldthat is uniform near the planet's surface. A 2.0 Kg ball with acharge of 5.0 C is thrown upward at a speed of 20.1 m/s. Ithits the ground after an interval of 4.1 sec. What is the potentialdifference between the starting point and the top point of thetrajectory? NOTE: The Cramster solution is not even close to the correctanswer, so only answer if you really know your Physics. Not a textproblem. I know that youn can use kinematics and V= QLD tosolve.

Explanation / Answer

I like the stipulation "only answer if you really know yourphysics". I hope I qualify, but I'll let you be thejudge. . (quick note... many of the cramster "solutions" are junk. Idont bother to write solutions any more because cramster ignoresmine. I dont know why.) . Anyway... there are two forces acting on the ball:  mg down   and   qE in downward directionalso. . Hmmm... I see a problem. Did you get this problem from anonline homework system? If so, you might be the victim of "numberrandomization" by the system. Why? Well, if there was no Efield it would take gravity alone .      v / g = 20.1 / 9.8 =    2.05 seconds...    to bring theball to a stop, and another 2.05 seconds to bring it backdown. . In other words, it would take   4.1 secondswith gravity alone. Which means in your case,   E fieldis zero (this happens sometimes with random numbergeneration...) . Let's suppose for a moment that you were given some time otherthan 4.1. Call it time "t". . Then we can write this: .       sum of forces = ma .       - mg - qE   = ma          a = - ( g + qE/m) . Now using kinematics: .     final position = initialposition   +    initial velocity *t   + (1/2) a t2 . In your case...       0   =   0    +    v t    - (1/2) ( g +qE/m) t2 . Or           ( g+ qE/m) t2 = 2 vt .       qE/m =  2v/t - g .      E   =   m (2v/t   - g) / q = 2.0 * ( 2 * 20.1 /4.1   - 9.8) / 5 x 10-6 = .              =   2.0 * ( 9.80   - 9.80 ) / 5 x 10-6    =    0 . If your time was anything but 4.1 seconds, youwould get a non-zero answer. . You could then calculate the max height reached by the ballusing: .       v2 = 2gh       h = v2 /2g    . and then the potential difference is just E field timesh. . Let me know if you have any questions, or if maybe you typoedthe numbers... . Or           ( g+ qE/m) t2 = 2 vt .       qE/m =  2v/t - g .      E   =   m (2v/t   - g) / q = 2.0 * ( 2 * 20.1 /4.1   - 9.8) / 5 x 10-6 = .              =   2.0 * ( 9.80   - 9.80 ) / 5 x 10-6    =    0 . If your time was anything but 4.1 seconds, youwould get a non-zero answer. . You could then calculate the max height reached by the ballusing: .       v2 = 2gh       h = v2 /2g    . and then the potential difference is just E field timesh. . Let me know if you have any questions, or if maybe you typoedthe numbers...

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