Problem 7.55 Part A A 177 g block is launc hed by compressing a spring of consta

Question

Problem 7.55 Part A A 177 g block is launc hed 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 "s After launch, where does the block finally come to rest? Measure from the left end of the fric tional zone Express your answer using two significant figures. 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 nse, as shown in the figure (Eigure 1 can Figure Submit My Answers Give Up of 1 Frictionless "0.27 Frictionless

Explanation / Answer

We can solve this question by simple work energy theorem

initially compressed spring contions potential energy , this energy will be used in work against friction on friction surface.

Let the stopping distance of block be L on the friction surface

Spring constant K = 200 N/m

Mass of ball = 177 g = 0.177 kg

Friction coefficient = 0.27

Compression in spring = 15 cm = 0.15 m

InitiInitial spring potential energy = 1/2 ( Kx2) = 0.5×200×(0.15)2 = 2.25 joule

Frictinal fforce = 0.27×0.177×9.8 = 0.468 N

Work done against for. L length = 0.468×L

Equating both the equation we get

0.468L = 2.25

Or. L = 4.81 m

But we have 85 cm of friction surface only.

So block will return from curve and again reapeat the motion until it loses all its potential energy to friction .

We can write L = 5×(0.85) + 0.56 that is , the block will cross the length

5 times and further 56 before coming to rest.

Hence distance of final position from left = 85-56 = 29 cm

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