Find the coefficient of friction between block and surface

AI Thread Summary
To find the coefficient of friction between a block and a surface, energy conservation principles are applied, equating initial kinetic energy to the energy stored in the spring and the energy lost to friction. The equations provided include kinetic energy (KE) and spring potential energy, which are essential for the calculations. The block, with a mass of 2.4 kg and an initial speed of 4.1 m/s, compresses a spring with a stiffness constant of 714.3 N/m by 3.7 cm before stopping. The discussion highlights the importance of correctly applying the equations and addressing any mistakes in the calculations. Proper formatting of equations is also emphasized for clarity.
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?
 
shivu30198 said:
umgx= mv2/2-kx2/2
u= mv2-kx2/2mgx
Some mistakes in that last step.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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