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qz3sol_3705sp10

qz3sol_3705sp10 - K of the block as a function of position...

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TA: Tomoyuki Nakayama Monday, February 8 th , 2010 PHY 2048: Physic 1, Discussion Section 3705 Quiz 3 (Homework Set #5) 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. ________________________________________________________________________________ In the figure below right, a block of mass m lies on a horizontal frictionless surface and is attached to one end of a horizontal spring (spring constant k ) whose other end is fixed. The block is initially at rest at the position where the spring is unstretched ( x = 0) when a constant horizontal force F in the positive direction of the x axis is applied to it. A plot of the resulting kinetic energy is shown below. The scale of its vertical axis is set by K s = 8.00 J. a) Write down the kinetic energy
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Unformatted text preview: K of the block as a function of position x . The block is initially at rest, thus the initial kinetic energy of the block is zero. Taking the origin at the relaxed position, the works done by the spring and the applied force are -(1/2)kx 2 and Fx, respectively. Applying the work-kinetic energy, we get ΔK = W K = -(1/2)kx ⇒ 2 + Fx b) What is the magnitude of the force F ? We plug the values of graph at x = 1 m and x = 2 m in the equation in part a), and solve them for F. 8 = - (1/2)k × 1 2 + F × 1 ⇒ 16 = - k + 2F 0 = (1/2)k × 2 2 + F ×2 ⇒ 0 = -2k+2F From the second equation, we get k = F. Plugging this into the first equation, we have 16 = - F + 2F ⇒ F = 16.0 N c) What is the spring constant k ? We already know that k = F from part b). Therefore, k = 16.0 N/m...
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