Hi, all A 3.0 kg block is released from acompressed spring (k=120 N/m). It trave
ID: 1727292 • Letter: H
Question
Hi, all A 3.0 kg block is released from acompressed spring (k=120 N/m). It travels over a horizontalsurface (mu =0.20) for a distance of 2.0 m before coming torest, Fig 1. How far was the spring compressed before beingreleased ? (Ans: 0.44m.) I need the fullanswer 'cause I tried this equ Wa =Ws (k x - m g mu ) d = 0.5 kx^2 but it doesn'twork tell me what wrong? Hi, all A 3.0 kg block is released from acompressed spring (k=120 N/m). It travels over a horizontalsurface (mu =0.20) for a distance of 2.0 m before coming torest, Fig 1. How far was the spring compressed before beingreleased ? (Ans: 0.44m.) I need the fullanswer 'cause I tried this equ Wa =Ws (k x - m g mu ) d = 0.5 kx^2 but it doesn'twork tell me what wrong?Explanation / Answer
Given that the mass of block is m = 3.0 kg Spring constant is K = 120 N/m The distance traveled before coming to rest is S = 2.0m Coefficient of kinetic friction is = 0.20 -------------------------------------------------------------------- From work energy theorem Work done by spring force = work done by the net force (1/2)Kx2 = (mg)S ( since there is only forceis frictional force ) x = ( 2mgS / K )1/2 = 0.44 m
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