Today we considered a spring pulling a mass along the table. There is no friction. The free-body diagram looks like:

Since the acceleration has no vertical component, then n = mg.

The spring force Fs has magnitude ks, where s is the stretch of the spring. From Newton's second law:

ks = ma

and thus

k = ma/s

Now we put in the numbers:

k = (0.5 kg) (1 m/s^{2}) /(1 cm)

= 50 kg· m/s^{2}

= 50 N/m

Notice that the units are consistent with Hooke's law: F = ks. Also notice that since we worked with force magnitudes, we did not need any minus signs.