Answer:
a) 0.24 J
b) 2.1 cm
Explanation:
The work needed to stretch a spring is given by:
[tex]W_s=\frac{1}{2}k*x^2[/tex]
where x is the elongation:
[tex]x=|x_f-x_i|\\x=46cm=36cm=10cm[/tex]
We need to know the spring constant so we can calculate what is asked in question a.
[tex]6J=\frac{1}{2}*k*(10*10^{-2}m)\\K=1200N/m[/tex]
so:
[tex]x=42cm-40cm=2cm[/tex]
[tex]W_s=\frac{1}{2}*1200N/m*(2*10^{-2}m)^2\\W_s=0.24J[/tex]
For b:
According to Hooke's law:
[tex]F=-k*x[/tex]
A force of 25N is pulling on the spring, so the spring pulls back with the same magnitude but in the opposite direction.
[tex]x=\frac{25N}{1200N/m}\\\\x=2.1cm[/tex]
