PHYS1131finalS12010SOL2.pdf

Now wcd pd vcd sincepdisfixedalongthefinalchange

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Unformatted text preview: ΔE AC = 800J, W ABC = −500 J = W AB + W BC € From conservation of energy ΔE ABC = ΔE AC since the same initial and final states. € Q + W from the 1st Law. = ABC ABC € So, QABC = ΔE AC − W ABC = 800 J − (−500 J ) = 1,300 J € (ii) We have PA = 5 PC , ΔVAB = −ΔVCD from inspection of the PV‐diagram. € Now WCD = −PD ΔVCD since PD is fixed along the final change. € We also know that W ABC = W AB + W BC = −500 J since W BC = 0 (V fixed) . € ∴ W = −P ΔV CD D CD + PA −W AB € = ΔVAB = = −(−500 /5) = +100 J 5 5 (iii) From the 1st Law ΔE int = Q + W € ΔE AC + ΔE CD + ΔE DA = 0 (cyclic) € We know that with ΔE AC = +800J (given) € So that ΔE CDA = −800 J = WCD since W DA = 0 (Volume fixed) W Also CDA Thus QCDA = ΔE CDA − WCD = −800 J − 100 J = −900 J € This is heat taken from the system. Thus +900J is added as heat to the surroundings. (iv) We know that ΔE DA = +500 J Since also ΔE DA + ΔE AC + ΔE CD = 0 (cyclic) Then 500 + 800 + ΔE CD = 0 € So that ΔE CD = −1,300 J ΔE = QCD + WCD (1st Law) Also, CD ∴ QCD = −1,300 − 100 J = −1,400 J € € Higher Physics 1A Q4 (a) PHYS1131 2010­T1, Solutions 12 Marks x Block A executes SHM given by ˙˙ = −ω 2 x with ω = 2πf and f=1.50 Hz. So maximum acceleration is given ω2xmax € Friction force between A&B given by F = µs M B g . Thus MB x Max. Acceleration ≡ Frictional Force before slip occurs 2 € ∴ M B (2πf ) x max = µs M B g µs g 0.6 × 9.8 i.e. x max = = = 0.0663m = 6.6cm to 2SF 4 π 2 f 2 4 π 21.5 2 (b) With Block C, μs=0.5 Maximum force for friction when µsC...
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This note was uploaded on 12/02/2012 for the course PHYS 1131 taught by Professor Angstman during the Three '12 term at University of New South Wales.

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