2/4 points | Previous Answers CJ10 10.P.040 My Notes Ask Your T A 1.4-kg object

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2/4 points | Previous Answers CJ10 10.P.040 My Notes Ask Your T A 1.4-kg object is suspended from a vertical spring whose spring constant is 135 N/m. (a) Find the amount by which the spring is stretched from its unstrained length. 0.1016 (b) The object is pulled straight down by an additional distance of 0.12 m and released from rest. Find the speed with which the object passes through its original position on the way up. 2.16 x m/s Additional Materials section 10.3 Submit Answer Save Progress Practice Another Version

Explanation / Answer

(a) By F = -kx {-ve just indicating the direction}
=>x= (mg)/k = (1.4 x 9.8)/135 = 0.10162

(b) By the law of energy conservation:-
=>1/2mv^2 = 1/2kA^2
=>1.4 x v^2 = 135 x (0.12)^2
=>v = 1.3885
=>v = 1.178 m/s

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