100 mm the spring is unstretched When the system is in equil

100 mm, the spring is unstretched. When the system is in equilibrium, d = 300 mm. The masses m1 = 10 kg

and m2 = 5 kg. Determine the spring constant k.

Solution

When the system is in equilibrium, the tension in the rope is
T = 5 kg*9.8 m/s^2 + k*(0.2 m),
where \"k\" is the spring constant.
The vertical component of this tension is 10 kg*9.8 m/s^2, holding up mass m1. At that position, the angle the rope makes with the horizontal is arctan(300/250), so you have:
10 kg*9.8 m/s^2 = T sin(arctan(6/5))
= T * 6/sqrt(25+36).
Hence,
T = 127.57 N, and
k*(0.2 m) = 127.57 N - 49 N = 78.57 N,
so the spring constant \"k\" is 393 N/m.