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 Rana_4778 #1 Posted : Monday, July 13, 2020 2:29:00 AM Rank: NewbieGroups: Registered Joined: 5/23/2019Posts: 1Thanks: 0 timesWas thanked: 0 time(s) in 0 post(s) I just wanted to know why the method I used was wrong and got me the wrong answer compared to the answer at the back of the book.Q. 92: A particle with +2C charge and 1 g mass is exposed to an electric field with strength 5N/C. How far will the particle move in 10s.CORRECT ANS: 5x10^5 mMy ANS: U = KEqEd = (1/2) mv^2qEd = (1/2) m(d/t)^2(2)(2C)(5N/C)d= (10^-3 kg)(d^2)/(t^2)(2)(2C)(5N/C)d= (10^-3 kg)(d^2)/(10^2)(20N)(10^2)d/(10^-3kg) = (d^2)(20N)(10^2)/(10^-3kg) = d2x10^5 m = dThis is the wrong answer.ANSWER AT THE BACK OF BOOK:F=qE=(2C)(5N/C)=10NF=ma ----> a=F/m = 10N/(10^-3kg) = 10^4m/s^2Distance traveled, x= (1/2)at^2 = (1/2)(10^4)(10)^2 = 5x10^5 mBasically, my question is why is my method wrong and leads to the wrong answer? And how do I know when to use energies versus using the ideas of the answer at the back of the book? Back to top User Profile
 INSTR_Katerina_102 #2 Posted : Tuesday, July 14, 2020 1:47:12 AM Rank: Advanced MemberGroups: Registered Joined: 6/24/2019Posts: 250Thanks: 0 timesWas thanked: 0 time(s) in 0 post(s) Hi,1.I think your method is flawed because it assumes that your particle is not accelerating in in the electric field.Because the particle is fully accelerating and not moving at constant velocity in an electric field, you have underestimated the distance that it would have covered. I also never really used KE to solve for distance on my MCAT, just for velocity.2.For this type of question, when I'm given 2 or 3 of following variables: v, vo, a, d, or t, and then asked to solve for another one of these. I often think of kinematics formulas first.Here, you are indirectly given a through the electric field, and you are asked for d after a given time t.What I found really helped me with these types of questions was just listing out what variables are given at first and then thinking of what formulas use the most of the variables in the question.I hope this helps! Back to top User Profile
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