Question

A light spring obeys Hooke's law. The spring's unstretched length is 32.0 cm. One end of the spring is attached to the top of a doorframe and a weight with mass 8.00 kg is hung from the other end. The final length of the spring is 45.0 cm. The load and the spring are taken down. Two people pull in opposite directions on the ends of the spring, each with a force of 180 N. Find the length of the spring in this situation.

Answer 1

Answer:

91.6cm

Explanation:

so F=kx

initially  f= 9.81 x 8kg

and x= the extension of spring so = 0.45-0.32= 0.13

k= f/x

k=(9.81x8)/0.13 = 603.7

force by both people pulling = 180+180= 360N

so again F=Kx (k dosent change from before as is fundamental property of spring)

360=603.7x

x=360/603.7= 0.596m

so final length of spring= 0.32+0.596=  0.916m

or 91.6cm :) !!

Answer 2

Answer:

$l"=61.82\ m$

Explanation:

Given:

length of unstretchered spring, $l=0.32\ m$

mass attached to the spring, $m=8\ kg$

stretched length of the spring, $l'=0.45\ m$

force on the spring ends, $F=180\ N$

Now we find the stiffness of the spring from the given case:

we know according to Hooke's law

$F=k.\delta l$ ......................(1)

where:

$F=$ force on the spring

k= stiffness constant

$\delta l=$ change in length of the spring due to load

Equation 1 becomes:

$m.g=k.(l'-l)$

$8\times 9.81=k\times (0.45-0.32)$

$k=603.6923\ N.m^{-1}$

Now the extension in the length of the spring:

$F=k.\Delta l$

$180=603.6923\times \Delta l$

$\Delta l=0.2982\ m=29.82\ cm$

Now the final length of the spring:

$l"=l+\Delta l$

$l"=32+29.82$

$l"=61.82\ m$