qz5sol_3885f10

# qz5sol_3885f10 - that work done by a force is given by the...

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TA: Tomoyuki Nakayama Monday, October 4, 2010 PHY 2048: Physic 1, Discussion Section 3885 Quiz 5 (Homework Set #6) Name: UFID: Formula sheets are not allowed. Calculators are allowed. Do not store equations in your calculator. You need to show all of your work for full credit. ________________________________________________________________________________ The only force acting on a 0.800 kg body as the body moves along an x axis varies as shown in the figure below right. The scale of the figures’ vertical axis is set by F s = 4.00 N. The velocity of the body at x = 0 is 6.00 m/s. a) At what value of x will the body have a kinetic energy of 10.0 J? The kinetic energy of the body at x = 0 is K i = (1/2)mv i 2 = 14.4 J In the region between x = 0 and 2 m, the force does some positive work and then some negative work. By symmetry, the total work is zero in this region. (Recall
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Unformatted text preview: that work done by a force is given by the area under the force vs position graph, with signs included.) This means the kinetic energy increases first and then decreases to the original value. Thus the kinetic energy of the body is always larger than 14.4 J. In the region between x = 2 m and 5 m, the kinetic energy of the body is expresses as K = 14.4 + (-F s )(x - 2) Setting the energy to be 10 J, and we solve the equation for x. 10 = 14.4 – F s (x – 2) ⇒ x = (14.4 – 10)/F s + 2 = 3.1 m b) What is the maximum kinetic energy of the body between x = 0 and x = 5.00 m? The kinetic energy of the body increases between x = 0 and 1 m and then decreases. Therefore, the kinetic energy takes the maximum value at x = 1 m. The maximum energy is K max = K (x = 1 m) = K i + (1/2)F s ×1 = 16.4 J...
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## This note was uploaded on 01/25/2011 for the course PHY 2048 taught by Professor Field during the Summer '08 term at University of Florida.

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