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If 4. 0 j of work are performed in stretching an ideal spring with a spring constant (force constant) of 2500 n/m, by what distance is the spring stretched?

Sagot :

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.

How to calculate the work done required stretching the spring?

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

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