A 180 g block is launched by compressing a spring of constant k =200N/m a distan

Question

A 180 g block is launched by compressing a spring of constant k=200N/m a distance of 15 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 ?=0.27. This frictional surface extends 85 cm, followed by a frictionless curved rise, as shown in the figure.(Figure 1)

After launch, where does the block finally come to rest? Measure from the left end of the frictional zone.

Explanation / Answer

K.E. of block = 0.5*mv^2 = 0.5kx^2
v = sqrt(200*0.15^2/0.18) = 5m/s

friction, f = 0.18*0.27*9.8 = 0.47628
deceleration = 2.646
2*2.646*s = 25, s = 4.724 m, which is more than the friction al surface, so the block would cover the 85 cm of frictional surface completely

velocity at the end of the frictional surface
2*2.646*0.85 = 25 - v^2
V = 4.527 m/s

Then the block would go up the curved rise, and the hieght it attains would be given by : mgh = 0.5mv^2
h = 1.046 m

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