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## Homework Statement

"A spring of negligible mass and spring constant 60 N/m is attached to a block of mass 5 kg, which rides on a horizontal surface with a frictional kinetic coefficient of 0.15. A string then goes over a perfect pulley ( no friction or moment of inertia) and attaches to a hanging block of mass 2 kg. Initially, the spring is at the unstretched and uncompressed position, and the blocks are released at rest."

Ok... they first want me to find the a) fricitional force acting on the 5 kg block as slides to the right

and b) then the maximum extension of the spring

## The Attempt at a Solution

Ok.. What I really want is for someone to agree with the logic that I will state below in trying to solve these problems and if not, please correct me.

a) Ok.... So the only way I think I can solve this is through Netwon's second law right?

So for Block attached to spring, Fy= Fn-m1g=0

and Fx=T-f=m1a? Right?... if we consider the axes as they regularly are?

For Hanging Block.... Fx=m2g-T=m2a... if we treat the x as positive pointing down...

.... but if this is solve in this way, then how do I find the acceleration?... I mean I can cancel the two tensions.... Fx=m2g-f=(m1+m2)a... but how do I get the acceleration...

Or is it that simple... that I just do Fn=m1g and mutiple the frictional kinetic coeffiecient?

Or the second part...

I know we have to use the conservation of energy....

So Ei=Ef

That means, initially there is no kinetic energy or potentinal energy at rest

so... Ei=0???

so... Ef= 1/2kx^2+1/2(m1+m2)v^2+m2g(-y)+f(-y)=0..Right????

I mean the spring has potential energy, then the two blocks are moving at the same velocity as a system, so they have kinetic energy, and the hanging block moves down, so it has potential energy, and the frictional force disippates energy in the form of thermal energy? Is this right... and if so.... how do we factor in the unknown factors of the velocity of the system, and the distance the two blocks move?

Thank you.