Version 001 – Energy Practice – Ross – (20691) 1 This print-out ...

Version 001 – Energy Practice – Ross – (20691) 9029 (part 6 of 6) 0.0 pointsFind the range ∆x of the projectile.Correct answer: 14.2524 m.Explanation:y = 0, soSincey = y i +v yi t− 1 2 gt2( g2v 2 x i)a = g2v 2 x i== y i + v y iv xix− g 2x 2 −(vyiv xi)x 2v 2 x i= 0x+y i = 09.8 m/s22(6.4 m/s) 2 = 0.119629 m−1 ,b = − v y i7.90334 m/s= −v xi 6.4 m/sc = −y i = −6.7 m, andb 2 −4ac = (−1.2349) 2= −1.2349,−4(0.119629 m −1 )(−6.7 m)= 4.73103,x = −b±√ b 2 −4ac2a= 1.2349±√ 4.731032(0.119629 m −1 )= 14.2524 m .Inclined plane consv energy030 0.0 pointsA block is released from rest, at a height h,and allowed to slide down an inclined plane.There is friction on the plane. At the bottomof the plane, there is a spring that the blockwill compress. After compressing the spring,theblockwillslideuptheplanetosomemaximumheight,h A ,afterwhichitwillagainslideback down.How much work is done on the block betweenits release at height h and its ascent toits next maximum height?1. 0 correct2. 2µm A gh A3. m A g(h−h A )+2µm A gh A4. More information is needed.5. None of these6. m A g(h − h A )Explanation:The important thing to remember here isthe work energy theorem, ∆K = ∆W. Sincethe block starts from rest and again is at restat its maximum height, then we know thekineticenergy is thesame, 0. Since the ∆K is0, then so must the ∆W. Therefore, no workis done.Pulled Block 02031 (part 1 of 2) 0.0 pointsGiven: A block is pulled at a constant speedalong a horizontal surface by a rope with tensionFT . The ropeisinclinedat ananglefromthe horizontal as shown in the figure below.The block moves a distance 15 m along thehorizontal surface.The acceleration of gravity is 9.8 m/s 2 .25 kgF T28 ◦µ = 0.7What is the net work done by all the forcesacting on the block?1. W net < 02. W net > 03. W net = 0 correctExplanation: