8 A 3.7-kg block is released from rest and allowed to slide down a frictionless

Question

8

A 3.7-kg block is released from rest and allowed to slide down a frictionless surface and into a spring. The far end of the spring is attached to a wall (see figure for problem 69 on page 231). The initial height of the block is 0.5 m above the lowest part of the slide. The spring compresses from the initial length of 49.3 cm down to the maximum compression of 10.7 cm.

What is the spring constant of the spring?

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CHEGG,

Please start with (show how you derive from) simple/basic formulas, and please clearly label and show ALL your work. Thank you!

Explanation / Answer

apply from the law of conservation of energy

Elastic potential Energy = Gravitation potential energy

0.5 kx ^2 = mgh

m is masss

g is accleration due to gravity

h is hieght

k is spring constant = ?

x is compresion = 49.3 - 10.7 = 38.6 cm

so

0.5 * K * 0.386^2 = 3.7 * 9.81 * 0.5

K = spring constant = 2* 3.7 * 9.81 * 0.5./(0.386^2)

K = 243.6 N/m

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