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qz4sol_6942s11 - The change in thermal energy of the...

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TA: Tomoyuki Nakayama Monday, February 21, 2011 PHY 2048: Physic 1, Discussion Section 6942 Quiz 4 (Homework Set #6) 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. ________________________________________________________________________________ A horizontal force of magnitude 50.0 N pushes a block of mass 4.00 kg across a floor where the coefficient of kinetic friction is 0.300. a) When the block slides through a displacement of 2.00 m, the thermal energy of the block increases by 15.0 J. What is the increase in thermal energy of the floor?
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Unformatted text preview: The change in thermal energy of the block-floor system is equal to the work done by the friction between the block and floor. We get ∆ E th = -W f = μ k mgd = 23.52 J The thermal energy goes to the floor, or to the block. Hence the change in thermal energy of the floor is ∆ E th = ∆ E th,B + ∆ E th,F ⇒ ∆ E th,B = ∆ E th - ∆ E th,F = 8.52 J b) During the displacement of 2.00 m, what is the increase in the kinetic energy of the block? According to the work-energy theorem, change in the kinetic energy of an object is equal to the work done by the net force exerted on the object. Only the applied horizontal force and friction do work on the block. The work-energy theorem yields ∆ K = W net = W F + W f = Fd - μ k mgd = 76.5 J...
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