The distance by which the spring is stretched when 4. 0 j of work are performed in stretching it with a spring constant (force constant) of 2500 n/m is 5.7 cm.
Work done of spring is the product of average force and the displacement. It can be given as,
[tex]W=\dfrac{1}{2}kx^2[/tex]
Here, k is the spring constant and x is the displacement.
The 4. 0 j of work are performed in stretching an ideal spring with a spring constant (force constant) of 2500 n/m. Thus,
[tex]W=4\text{ J}\\k=25000\text{ N/m}\\[/tex]
Put the values in the formula.
[tex]4=\dfrac{1}{2}(2500)(x)^2\\x^2=\dfrac{4\times2}{2500}\\x=\sqrt{0.0032}\\x=0.057 \text{m}\\x=5.7 \text{cm}[/tex]
Thus, the distance by which the spring is stretched when 4. 0 j of work are performed in stretching it with a spring constant (force constant) of 2500 n/m is 5.7 cm.
Learn more about the work done of spring here;
https://brainly.com/question/3317535
#SPJ4