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ω = √( k /m) = √(5kg/s² / 0.25kg) = 4.47 rad/s Measured from the new equilibrium...

 the equation: k = – F/x. Where k is the spring constant, F is the force applied over x, and x is the displacement by the spring F =mg Subbing in we get k = mg/x N/m So make sure to convert cm to m

friction = µ k * normal force f = µ k * (mg +Fy) = 0.9*(12kg*9.8m/s² + 40N) = 142 N acceleration = net force / mass a = (150N - 142N) / 12kg = 0.68 m/s²

PE = .5kx^2 where k is a constant and x is the length stretched PE= mgh...

...difference must be absorbed by the spring. then use E = ½kd²   k is spring constant in N/m you are missing k , but you need...

... from ma = 7*3 = 21N  But the net force = F - mg*μ k   -> F = 21 + 7*9.8 * 0.2  ~= 34.7N To accelerate the small block...

By kinematics: acceleration = net force / mass = -µ k *m*g / m = -µ k *g a = -0.15 * 9.8m/s² = -1.47 m/s² ...

frequency f = (1/2π)*√( k /m) + (1/2π)*√(7600kg/s² / 105kg) f = 1.35 Hz ◄ period T = 1/f = 0.739 s ◄ Hope this helps!