sol6a - HW Solutions # 6 - 8.01 MIT - Prof. Kowalski...

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HW Solutions # 6 - 8.01 MIT - Prof. Kowalski Potential Energy and Mechanical Energy Conservation. NOTE: We made a mistake on the solution sheet for HW #5. On part d of problem 5 on HW #5 we omitted the work due to friction. If you submitted the answer 1 2 mv 2 0 x 2 x 1 x 1 + 1 2 mv 2 1 it should have been graded correct , and you can recover any lost credit by resubmitting it to your grader with a note to regrade this problem. 1) 7.42 Please refer to the figure 7.30 p.277. Let’s denote the compression length as ∆ x , the velocity after leaving the spring as v 0 and the maximum height it goes on the incline as h . Using energy conservation law: 1 2 mv 2 1 + 1 2 kx 2 1 + mgy 1 + W other = 1 2 mv 2 2 + 1 2 2 2 + mgy 2 (1) There is no friction present in this problem and W other = 0. I will measure the gravitation potential energy with respect to the height of the dashed horizontal bar indicated in the Fgure 7.30. a) Comparing the energy initially (when spring compressed): v 1 =0 x 1 =∆ xy 1 (2) with the point it leaves the spring: v 2 = v 0 x 2 y 2 (3) 0+ 1 2 k x 2 +0= 1 2 mv 2 0 +0+0 (4) 1
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v 0 = ± k m x (5) b) For this part we should compare the energy of particle at its highest point on the incline: v 3 =0 x 3 y 3 = h (6) with the point that it has left the spring: v 2 = v 0 x 2 y 2 (7)
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sol6a - HW Solutions # 6 - 8.01 MIT - Prof. Kowalski...

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