FALL_2013_PHY1321_assignment5_solutions

# 250 400 n 100 n f x emech 100 200 u a ub

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Unformatted text preview: f − hi ) = ( 50.0) ( 9.80) ( 20.0 sin 37.0°) = 5.90 × 10 3 ( ) ΔUB = mB g h f − hi = (100) ( 9.80) ( −20.0) = −1.96 × 10 4 1 mA v 2 − vi2 f 2 m 1 ΔKB = mB v 2 − vi2 = B ΔK A = 2ΔK A f 2 mA ΔK A = ( ( ) ) Adding and solving, ΔKA = 3.92 kJ . 3. A skier of mass 70.0 kg is pulled up a slope by a motor- driven cable. (a) How much work is required to pull him a distance of 60.0 m up a 30.0° slope (assumed frictionless) at a constant speed of 2.00 m/s? (b) A motor of what power is required to perform this task? (a) ∑ W = ΔK , but ΔK = 0 because he moves at constant speed. The skier rises a vertical distance of ( 60.0 m) sin 30.0° = 30.0 m . Thus, ( ) Win = −Wg = (70.0 kg ) 9.8 m s2 ( 30.0 m) = 2.06 × 10 4 J = 20.6 kJ . (b)...
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