oblem 7.55 Part A A 181 g block is launched by compressing a spring of After lau

Question

oblem 7.55 Part A A 181 g block is launched by compressing a spring of After launch, where does the block finally come to rest? Measure from the left end of the frictional constant k-200N m a distance of 15 cm. The spring zone. s mounted horizontally, and the surface directly under it s frictionless But beyond the equilbrium position of the spring end, the surface has coefficient of friction = 0.27 This frictonal surface extends 85 cm. followed by a frictionless curved rise, as shown in the s your answer using two significant figures. AE Figure1 of 1 CID Submit My Answers Give Up Frictionless =027 Frictionless

Explanation / Answer

The kinetic energy of the block = 0.5*181*0.15^2 = 2.036 J
The frictional force, f acting on the block = 0.181*9.8*0.27 = 0.479 N.

The work done in crossing 85 cm of frictional path = 0.85f= 0.407 J.

So the kinetic energy of the block at the beginning of frictionless rise = 2.036 - 0.407 = 1.629 = 0.181*9.8*h

or

h = 0.918 m. So the block rises to a height of 91.8 cm

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