A metal cube suspended freely from the end of a spring causes it to stretch by 5 cm. A 500g mass suspended from the same...

      

A metal cube suspended freely from the end of a spring causes it to stretch by 5 cm. A 500g mass suspended from the same spring stretches it by 2 cm. If the elastic limit is not exceeded, by what length will the spring stretch if a mass of 1.5 kg is attached to its end?

  

Answers


Jim
Hooke’s law; F=ke
So K=F/e
K=0.5x10/2x10^-2
=250N/m
Weight of cube=250x5x10^-2
=12.5N
Let the extension produced by the 1.5 kg mass be e.
F=mg
Force=1.5x10
=15N
From F=ke
E=F/k=15/250
0.06m 0r 6cm.

jim items answered the question on September 17, 2018 at 08:46


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