The boy determined the maximum force with which he can stretch the dynamometer. it turned out to be equal to 400N.

The boy determined the maximum force with which he can stretch the dynamometer. it turned out to be equal to 400N. What work is done by this force when the spring is stretched? The spring rate of the dynamometer is 100,000 N / m.

The work of the elastic force can be calculated using the formula:

A = k * ∆l2 / 2, where k is the stiffness of the spring (k = 100,000 N / m); ∆l – elongation, amount of deformation of the spring (m).

Elastic force:

Fcont. = k * ∆l, where Fcont. – elastic force (Fel. = 400 N).

Let’s calculate the amount of spring deformation:

∆l = Fcont. / k = 400/100000 = 0.004 m.

Let’s calculate the work of the elastic force:

A = k * ∆l2 / 2 = 100000 * 0.0042 / 2 = 0.8 J.

Answer: When the spring is stretched, a work of 0.8 J.