Question

A block of mass m = 1.91 kg attached to a horizontal spring with force constant k = 6.85 x 10' N/m that is secured to a wall is stretched a distance of 5.20 cm beyond the spring's relaxed position and released from rest. (a) What is the elastic potential energy of the block-spring system just before the block is released? 3 (b) What is the elastic potential energy of the block-spring system when the block passes through the spring's relaxed position? j (c) What is the speed of the block as it passes through the spring's relaxed position? m/s (d) What is the speed of the block when it has compressed the spring 2.60 cm beyond its relaxed position?

solve and show work

Answer 1

10 = 0 - 926 JP tx=520 ch mm mmm Given : m = 1.91 kg torce constant = K = 6.858 102N) m Displacement from meen positim sem = 5.20 cm xm=5.20x1o²m. Eleistic potential energy just before the block is released P. E u I kan ₂ x 6.85 (5.20 x 102) 2 x 102 U = £ x 6.85+) 02 x 27-04 x 154 2 185.224x102 92.612x102 2 dolicrona wode drove on (b) Elastic potential energy of spring block system when the black passes through the spring's related position when the block passes through the spring's relaxed position x=0 U = I ka? 2 1 x 6.85* )0² x 0 = 0 2 Loro -- U : 0 ols ten speed of the block as it passes through the spring's relareel position. maximum sp x=0, It hal speed At I'm2 = har det 2 62 = K. Xon 8.858 902 5.20 5 102)2 1:9) 185.224x102 1:9)

96.97 x 102 = 0.9697 0.9697 0.984 mis 0 = 0.984 mis speed of the block when it compressed the spring 2.60cm. x= - 2.60cm. £ muokkaa 2 t.22 6.858102 (-2.608102)2 D 1:9) 4 6.3068 102 = 24.24x102 1.9) 62 = 0.2424 Jo. 2424 = 0.492 mis of 0.5 mis