Conservation of Momentum and Mechanical Energy

[TG3]

Homework Statement

A projectile with a mass of 20 g has an initial horizontal velocity of 100 m/s when it hits and stops in a wood block of mass 0.402 kg. The block is sitting on a horizontal frictionless surface and is attached to a massless spring, initially relaxed, with spring constant 143 N/m. What is the maximum compression of the spring?

Homework Equations

P= MV
K=1/2 MV^2
F=-kx

The Attempt at a Solution

Initial Momentum of the Bullet = .02 x 100 = 2
Initial momentum of the Bullet/Wood system = 2 due to conservation of momentum.
2= .404 V
V= 4.95 m/s is the velocity of the block/bullet system.
K = 1/2 (.404) 4.95^2
K= 4.95 Due to conservation of Energy, this is also the amount of energy the spring exerts.
4.95 =143X
Wrong. I suspect the error in my calculation is near the end, but I don't know that for certain.

 

[LowlyPion]

20g is .02kg.
Total mass is .422 kg


Secondly the kinetic energy is right to use, but maybe consider the potential energy relationship with the spring instead of the force relationship?

http://hyperphysics.phy-astr.gsu.edu/hbase/pespr.html#pe2

 

[TG3]

Thanks- once I used energy not force it was easy to find the distance. (.26 meters.)
In the second part though, there is friction (with a coefficient of friction of .25).

How do I deal with that using the energy relationship?

 

[Delphi51]

KE of block = PE gained by spring + work done against friction