Question

Two equal masses are attached to the two ends of a spring of spring constant k. The masses are pulled out symmetrically to stretch the spring by a length x over its natural length. The work done by the spring on each mass is:

(a) (1/2)kx²
(b) kx²
(c) (1/4)kx²
(d) (1/8)kx²

Answer 1

The work done by the spring on each mass when stretched symmetrically is (1/4)kx², since each mass contributes to half of the energy stored in the spring.

Step-by-step explanation:

When two equal masses are pulled symmetrically to stretch a spring by a length x over its natural length, the work done by the spring on each mass can be calculated using the formula for the potential energy stored in the spring, which is W = (1/2)kx². The spring constant k directly influences this work. However, since two equal masses are attached to the spring, each mass is only responsible for half of the stretch, so the work calculated by the full stretch should be divided by two.

Thus, the work done by the spring on each mass is (1/4)kx², because the energy is equally divided between both sides of the spring. Substituting the given spring constant and displacement will give the work done on one of the masses.