A spring, of negligible mass and which obeys Hooke\'s Law, supports a mass M on

Question

A spring, of negligible mass and which obeys Hooke's Law, supports a mass M on an incline which has negligible friction. The figure below shows the system with mass M in its equilibrium position. The spring is attached to a fixed support at P. The spring in its relaxed state is also illustrated.

Mass M has a value of 205 g. Calculate k, the spring constant.

Now please explain how do you calculate the length of spring once it has been stretched. I know its sqrt(60^2+30^2). But how do I obtain this 30??? Also why is delta x=46???

Explanation / Answer

From the diagram, the unstretched spring has length 30 cm.
The stretched spring has length (60² + 30²) cm =67.08 cm.
x =37.08 cm
angle of incline: = arctan(30/60) = 26.56º (from spring geometry)
downslope component of weight Fg = mgsin = 0.205kg * 9.8m/s² * sin26.56º = .898 N
spring force = kx = Fg
k * 0.3708m = .898 N
k = 2.42 N/m

b) = (k / m) = (2.42kg/s² / 0.205kg) = 3.43 rad/s
f = / 2 = 0.545 Hz

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