Question

An object of mass m = 2.50 kg is travelling on a horizontal surface. The coefficient of kinetic friction between the object and the surface is µk = 0.160. The object has speed v = 1.39 m/s when it reaches x = 0 and encounters a spring. The object compresses the spring a distance d/2, stops instantaneously, and then travels back to x = 0 where it stops completely.

What is the spring constant, k, of the spring?

Answer 1

This is a wonderful problem. If you use kinetics it won't be very fun; so use energy instead! You don't even really need the second half of the problem...


\[ \ell = d/2
\\ v = 1.39 m/s
\\ m = 2.5 kg
\\ \mu_k = .160
\]

\[ E_i = E_f\]
The only energy you have in the beginning of the problem is kinetic energy; when the block stops while pushing the spring, it has zero kinetic energy, lots of spring potential energy, and lots of energy "lost" through work done on the block by friction.
\[ K = U_{spring} + W_F\]
\[\frac{1}{2}mv^2=\frac{1}{2}kx^2 + F_F \cdot \ell
\\ \frac{1}{2}mv^2 =\frac{1}{2}kx^2 + F_F \ell cos \theta\]
The force of friction and the distance over which it is exerted are along the same coordinate (x) so our dot product is cos 0 = 1, giving
\[\frac{1}{2}mv^2=\frac{1}{2}k\left( \frac{d}{2}\right)^2 + \frac{1}{2}F_F d \]

\[ F_F = N \mu_k \]
Simplify and solve for k. should be in units of kg/s^2. and positive.