A 187 g block is launched by compressing a spring of constant k=200{\ m N/m} a d

Question

A 187 g block is launched by compressing a spring of constant k=200{ m N/m} a distance of 15 { m cm}. The spring is mounted horizontally, and the surface directly under it is frictionless. But beyond the equilibrium position of the spring end, the surface has coefficient of friction mu = 0.27. This frictional surface extends 85 { m cm}, followed by a frictionless curved rise.
After launch, where does the block finally come to rest? Measure from the left end of the frictional zone.
Express your answer using two significant figures

Explanation / Answer

let the acceleration of the block be a we know that ma = kx or a = kx/m where k = 200 N/m,x = 15 cm = 15 * 10^-2 m and m = 187 g = 187 * 10^-3 kg let the initial speed of the block be u we know that (1/2)mu^2 = (1/2)kx^2 or mu^2 = kx^2 or u^2 = kx^2/m or u = (k/m)^1/2 * x the final speed of the block is zero that is v = 0 therefore v = u + a * t where t is time taken by block to come to rest or t = u/a the distance travelled by block before coming to rest is S = ut + (1/2)u_k * a * t^2 where u_k = 0.27

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