A 189g block is launched by compressing a spring of constant k= 200 N/m a distan

Question

A 189g block is launched by compressing a spring of constant k= 200 N/m a distance of 15cm . 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 85cm , followed by a frictionless curved rise, as shown in the figure. After launch, where does the block finally come to rest? Measure from the left end of the frictional zone. Give answer in cm.

Explanation / Answer

A 190 g block is launched by compressing a spring of constant k=200N/m a distance of 15cm. 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 85cm, 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. The energy of the block when it first encounters friction (at point O) is k0 = (1/2)* 200* 0.15^2 = 2.25 J Loss due to friction µ*mgl = - 0.27* 0.19*9.8*8.5 = -0.427 J Since k0 / [µ*mgl ] = 5.27 and since 5 crossings are made 2.25 - 5*(0.427) = 0.115 J on the curved side. This remaining energy is sufficient to move the block a distance s = 0.115 ÷ µmg = 22.9 cm towards point O, so the block comes to rest 85 - 22.9 = 62.1cm to the right of point O.

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