Analyzing Block & Spring Compression: F = -kx & W = -0.5kx^2

[Pavel1989]

A block of mass 0.528 kg slides with uniform velocity of 3.85 m/s on a horizontal frictionless surface. At some point, it strikes a horizontal spring in equilibrium. If the spring constant is k=26.7 N/m, by how much will the spring be compressed by the time the block comes to rest? What is the amount of compression if the surface is rough under the spring, with a coefficient of kinetic friction uk = 0.411?


F = -kx W = -0.5kx^2 KE = 0.5mv^2



I found that it has 3.913 J of energy once it hits the srping, but how do i use that and incorporate it into the work done by the spring?

 

[Astronuc]

The spring mechanical energy = work done by/on the spring.

Without dissipating force (e.g. friction), the kinetic energy of the mass is transformed into the mechanical (elastic) energy stored in the spring.

With friction present, some of the kinetic energy is dissipated by friction.

Think of work = $\int_0^x F(s) ds$, and there will be + or - sign depending whether work is done on or by the system (e.g. spring).

 

[Pavel1989]

thanks much i think I've got it