Question

A spring (oriented horizontally, k = 40 N/m) is attached to the left side wall in a room whose floor is frictionless. A small, dense mass (m = 0.5 kg), at rest on the floor, is attached to the right side of the spring. The unelongated spring length is 0.6 m. A person then begins to pull on the system to the right with an applied force of 20 N. This force is applied until the spring elongates by 0.25 m. The force then instantly disappears. Find the speed of the block when the applied force vanishes.

Answer 1

Let x be the distance to which the spring is stretched. Then the net force exerted on the mass is

$F(x) = 20\,\mathrm N - \left(40\dfrac{\rm N}{\rm m}\right) x$

Then the total work performed on the mass as it's stretched to 0.25 m from equilibrium position is

$\displaystyle \int_0^{0.25\,\rm m} \left(20\,\mathrm N - \left(40\dfrac{\rm N}{\rm m}\right)x\right) \,\mathrm dx = 3.75 \,\mathrm J$

By the work-energy theorem, the total work done on the mass is equal to its change in kinetic energy. The mass starts at rest and is accelerated by the 20 N force to a speed v such that

$W_{\rm total} = \Delta K \\\\ 3.75\,\mathrm J = \dfrac12(0.5\,\mathrm{kg})v^2$

Solve for v :

$3.75\,\mathrm J = \dfrac12(0.5\,\mathrm{kg}) v^2 \\\\ v^2 = \dfrac{3.75\,\rm J}{0.25\,\rm kg} \\\\ v = \sqrt{\dfrac{3.75\,\rm J}{0.25\,\rm kg}} \approx \boxed{3.87\dfrac{\rm m}{\rm s}}$