A spring of a natural length 3cm is extended by 100cm by a force of 9N,what will be its length when the applied force is10N.
Given:
"l_0=0.03\\:\\rm m"
"l_1=1.03\\:\\rm m"
"F_1=9\\:\\rm N"
"F_2=10\\:\\rm N"
The Hooke's law says
"F=k(l-l_0)"Hence
"l_2=l_0+\\frac{F_2}{F_1}(l_1-l_0)\\\\\n=0.03+\\frac{10}{9}*1.0=1.14\\:\\rm m=114\\: cm"
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