Question

A 2 kg mass rests on a horizontal table with a light spring with a force constant of 200 Nm-1

attached to it. The coefficient of static friction between the mass and the table is 0.7 and the
coefficient of kinetic friction is 0.5. How far must the spring be stretched horizontally before the
mass begins to move and what is its initial acceleration when it does begin to move?

Answer 1

Wm = m*g = 2kg * 9.8N/kg = 19.6 N. = Wt.
of the mass.

Fap-Fs = m*a
Fap-u*mg = m*0
Fap-0.7*19.6 = 0
Fap=13.72 N. = Force applied by spring.

d = (13.72N/200N) * 1m = 0.0686 m. =
distance the spring is stretched.

Fn = Fap - Fk = 13.72-0.5*19.6 = 3.92 N.
= Net Force.

a = Fn/m = 3.92/2 = 1.96 m/s^2.