qz3sol_3706Hs11 - While the block moves from x = 5 m to x 1...

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TA: Tomoyuki Nakayama Tuesday, February 15, 2011 PHY 2048: Physic 1, Discussion Section 3706H Quiz 3 (Homework Set #5) Name: UFID: Formula sheets are not allowed. Do not store equations in your calculator. You have to solve problems on your own; memorizing final algebraic expressions from homework assignments and just plugging numbers into them will not give you full credit. Leave all your work. ________________________________________________________________________________ The only force acting on a 2.00 kg body as the body moves along an x axis varies as shown in the figure below right. The scale of the figure's vertical axis is set by F s = 15.0 N. The velocity of the body at x = 5 m has a magnitude of 8.00 m/s and is pointed in the negative x direction. a) What is the kinetic energy of the body at x = 1.50 m? The kinetic energy at x = 5 m is K i = (1/2)mv 2 = 64 J
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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 work-energy theorem, we get K max = K i + W 1 + W 3 = 116.5 J...
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This note was uploaded on 12/10/2011 for the course PHY 2048 taught by Professor Field during the Spring '08 term at University of Florida.

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