Block launched from a spring across a surface with friction then up a slope

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A 170 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 mu = 0.27. This frictional surface extends 85 cm, 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.

Homework Equations

delta,x=(u^2)/2*mu*g

u0=1/2*k*x0^2

The Attempt at a Solution

I found u0 by using u0=1/2*k*x0^2 to be 2.25 and worked out delta,x to be 4.9967m which has me stumped as to what to do next? any help will be useful thanks

 

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Also i would like to add tht there is a similar problem in wolfson essential university physics volume 1 chapter 7 problem 57 pg 116 tht has a mass of 190 g instead of 170 g

 

[rishabhdutta]

Force exerted on the block= k*x=200*.15=30N
This force causes the block to move with an acceleration of 30/.17=176.47m/s2
Now the block moves on a non-frictional surface upto 15 cm and then on a frictional surface of 85 cm. Its motion on the latter surface is characterized by a deceleration of mu*g= 2.6m/s2.
1st surface: .15=.5*176.47*t*t implies t=0.04123
so velocity before it reaches second surface= .04123*176.47=7.276m/s
2nd surface: .85= 7.276*t - .5*2.6*t*t solve quadratic
t=.1196s
velocity at the end of 2nd surface= 7.276*.1196-2.6*.1196=.56m/s

the remaining you may figure it out yourself

 

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Thanks for your help :)

 

[rishabhdutta]

though a correction

velocity at the end of 2nd surface= 7.276-2.6*.1196=6.97m/s