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How much work is required to stretch an ideal spring of spring constant (force constant) 40 N/m from x

Sagot :

Cricetus

Answer:

The work done will be "0.45 J".

Explanation:

Given:

K = 40 N/m

x₁ = 0.20 m

x₂ = 0.25 m

Now,

The required work done will be:

= [tex]\frac{1}{2}k[x_2^2-x_1][/tex]

By putting the values, we get

= [tex]\frac{40}{2}[(0.25)^2-(0.20)^2][/tex]

= [tex]20\times 0.0225[/tex]

= [tex]0.45 \ J[/tex]

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