Unformatted text preview: While the block moves from x = 5 m to x 1 = 2 m, the work done on the body is W 1 = (F s )(x 1 – x ) = +45 J While the block moves from x 1 = 2 m to x 2 = 1.5 m, the work is W 2 = (3/4)F s (x 2 – x 1 ) = + 5.63 J The work energy theorem yields the kinetic energy at x = 1 m: K f = K i + W 1 + W 2 = 114.6 J b) What is the maximum kinetic energy of the body between x = 0 and x = 5.0 m? Work done on the body is positive between x = 5 m and x = 1m, and it is negative between x = 1 m and x = 0 m. Hence the kinetic energy is maximum at x = 1 m. We have already calculated the work done on the body when the body moves to from x = 5 m to x 1 = 2 m. The work done on the body when the body moves from x 1 = 2 m to x 3 = 1 m is W 3 = (1/2)F S (x 3 – x 1 ) = + 7.5 J Applying the workenergy theorem, we get K max = K i + W 1 + W 3 = 116.5 J...
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 Spring '08
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 Energy, Kinetic Energy, Potential Energy, TA, 0 m, Tomoyuki Nakayama

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