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qz4sol_3081sp10 - to the block we have ΔE = W f W T Since...

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TA: Tomoyuki Nakayama Monday, February 15 th , 2010 PHY 2048: Physic 1, Discussion Section 3081 Quiz 4 (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. ________________________________________________________________________________ A rope is used to pull a 4.00-kg block at constant speed 5.00 m along a horizontal floor. The force on the block from the rope is 10.0 N and directed 25.0º above the horizontal. a) What is the work done by the rope’s force? The definition of work due to a constant force gives W T = Tdcosθ = 45.3 J b) What is the increase in thermal energy of the block-floor system? Thermal energy is given by the negative of work done by friction. Applying the work-energy theorem
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Unformatted text preview: to the block, we have ΔE = W f + W T Since the speed of the block is constant, change in kinetic energy is zero; the gravitational potential energy does not change either because the block is moving on a horizontal surface. Therefore, the change in mechanical energy is zero. Thus we have 0 = W f + W T ΔE ⇒ th ≡ - W f = W T = 45.3 J c) What is the coefficient of kinetic friction between the block and floor? Taking +y axis vertically up, we apply Newton’s 2 nd law along the y-axis to get the normal force. 0 = N + Fsinθ – mg N = mg - Fsinθ = 35.0 N ⇒ Expressing the change in thermal energy in terms of normal force and displacement, we get ΔE th = - W f = +μ k Nd μ ⇒ k = ΔE th /Nd = 0.259...
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