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Transcript
Phys101 Term:131 Online HW-Ch07-Lec01 Q1: An object moves in a circle at constant speed. The work done by the centripetal force is zero because: A. B. C. D. E. Ans: the displacement for each revolution is zero the average force for each revolution is zero there is no friction the magnitude of the acceleration is zero the centripetal force is perpendicular to the velocity T 90ᵒ v �⃗ v and ∆x has the same direction E ; W = ∆x�⃗. T Q2: At time t = 0 a 2.0 kg particle has a velocity of (4m/s) - (3m/s)j. At t = 3s its velocity is (2m/s)i + (3m/s)j. During this time the work done on it was: Ans: v0 2 = 42 + (−3)2 = 16 + 9 = 25 v 2 = 22 + 32 = 4 + 9 = 13 1 1 1 1 W = ∆k = k − k 0 = mv 2 − mv0 2 = m(v 2 − v0 2 ) = × 2(13 − 25) 2 2 2 2 ∴ W = −12 J Q3: A 0.50-kg object moves on a horizontal circular track with a radius of 2.5 m. An external force of 3.0N, always tangent to the track, causes the object to speed up as it goes around. If it starts from rest its speed at the end of one revolution is: Ans: The work done by the force for small displacement dx is: dw = dxf Total work done W = � 2πr 0 Fdx = F � 2πr 0 dx = F 2πr r dx F ∴ W = 2πrF = 2 × 3.14 × 2.5 × 3 = 13.7 m/s KFUPM-Physics Department 1
