Find the coefficient of friction between block and surface

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SUMMARY

The discussion focuses on calculating the coefficient of friction between a block and a surface using energy conservation principles. A block with a mass of 2.4 kg, traveling at a speed of 4.1 m/s, compresses a spring with a stiffness constant of 714.3 N/m by 3.7 cm before coming to a stop. The relevant equations include the work done by friction and the energy stored in the spring, leading to the conclusion that the coefficient of friction can be determined through the relationship between kinetic energy and spring potential energy.

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  • Understanding of kinetic energy and potential energy concepts
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shivu30198
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Homework Statement
A block of mass 2.4 kg sliding along horizontal rough surface is traveling at a speed 4.1 m/s when strikes a massless spring and compresses spring a distance 3.7 cm before coming to stop. If the spring has stiffness constant 714.3 N/m, find coefficient of friction between block and surface.
Relevant Equations
Wfr= umgcos
Ff.x= mv2/2 - kx2/2
umgx= mv2/2-kx2/2
u= mv2-kx2/2mgx =
 
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shivu30198 said:
Homework Statement: A block of mass 2.4 kg sliding along horizontal rough surface is traveling at a speed 4.1 m/s when strikes a massless spring and compresses spring a distance 3.7 cm before coming to stop. If the spring has stiffness constant 714.3 N/m, find coefficient of friction between block and surface.
Relevant Equations: Wfr= umgcos

Ff.x= mv2/2 - kx2/2
umgx= mv2/2-kx2/2
u= mv2-kx2/2mgx =
What is your question?

Also: please see LaTeX Guide to properly format your equations.
 
Use energy conservation.
Since block comes to rest:
##(KE)_{initial}=\text{ Energy stored in spring }+\text{ Energy lost to friction } ##
 
But you seem to know that, so are you asking us to do the mathematics?
 
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Likes   Reactions: erobz
shivu30198 said:
umgx= mv2/2-kx2/2
u= mv2-kx2/2mgx
Some mistakes in that last step.
 
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