Question

An ideal spring is fixed at one end. A variable force F pulls on the spring. When the magnitude of F reaches a value of 49.8 N, the spring is stretched by 18.1 cm from its equilibrium length. Calculate the additional work required by F to stretch the spring by an additional 10.9 cm from that position.

Answer 1

Answer:

W = 7.06 J

Explanation:

From the given information the spring constant 'k' can be calculated using the Hooke's Law.

$F = kx\\49.8 = k(0.181)\\k = 275.13~N/m$

Now, using this spring constant the additional work required by F to stretch the spring can be found.

The work energy theorem tells us that the work done on the spring is equal to the change in the energy. Therefore,

$W = U_2 - U_1\\W = \frac{1}{2}kx_2^2 - \frac{1}{2}kx_1^2 = \frac{1}{2}(275.13)[0.29^2 - 0.18^2] = 7.06~J$