A bloc of 10 kg is put at the top of an inclined plan of 45 degrees (to the left), attached to a spring which has a spring constant of 250 N/m. The coefficient of kinetic friction between the bloc and the surface of the inclined plan is of 0,300.

Question

What is the maximum elongation of the spring?

I know that maximum elongation must mean that when the block has a velocity of 0 m/s. Are speeds applied to springs as well?

Answer 1

Huh?

the force down the plane is mg*sinTheta.

Now, the energy in the spring will be 1/2 kx^2,and the energy lost due to friction will be x*mg*cosTheta*mu

so force*x=1/2 kx^2=mgCosTheta*mu*x

that is , the work done by gravity equals the work done by friction+energyinspring

Answer 2

is it always sinTheta for a force down a plane? Or does it depend on the inclination (left, right)

Answer 3

and what does mu stand for in x*mg*cosTheta*mu ?

Thank you for your time!

Answer 4

mu stands for the coefficent of friction

Answer 5

Okay so from what I understand, I find the elongation from the total Work done on the spring and of course by isolating it in my equation. Sorry if my terms aren't good, I do it in french and try to translate it as good as possible