Small typo: In part 1 you are asked to calculate the kinetic energy at x =...

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А PARTICLE IS MoviNG ALONG THE X-AXIS IT STARTS. FROM REST AT X - AND MOVES TO X = L UNDER THE ACTION OF A VARIABLG FORCE X-C

Small typo:

In part 1 you are asked to calculate the kinetic energy at x = L, so it should say K(x= L) NOT K(x=L/2).
In part 2 you are asked to calculate the kinetic energy at x = 3L, so it should say K(x= 3L) NOT K(x=L).

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Answer 1

Answer #1

sol: Ę (x) = OLULL 2 [ _ _ Lx if we hL 3L A we know, work done = change of Kinetic energy Here at initial point particle has L = 1/3 [23] + 2 2 5 11 7 4 2 4 [ro] 7 / 20 x 24) - 7 / 4 (95720) mim was + 33 2 3 th 3. 4. 3 [w velocity C) AAN=36, Kinetic

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Answer 2

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