Final velocity of spring down incline with fricition

[gap0063]

Homework Statement

A spring with a spring-constant 2.5 N/cm is compressed 35 cm and released. The 4 kg mass skids down the frictional incline of height 37 cm and inclined at a 20◦ angle. The acceleration of gravity is 9.8 m/s² . The path is frictionless except for a distance of 0.5 m along the incline which has a coefficient of friction of 0.3.
What is the final velocity vf of the mass?
Answer in units of m/s.

k=2.5 N/cm= 250 N/m
d=35 cm=.35m
m= 4 kg
h=37cm=.37,
theta=20◦
distance of friction = .5 m
Coefficient of friction=.3

Homework Equations

W_nc=[tex]\Delta[/tex]K+[tex]\Delta[/tex]U_spring+[tex]\Delta[/tex]U_gravity

The Attempt at a Solution

v_i=0
so W_nc=K_f+ [tex]\Delta[/tex]U_spring+[tex]\Delta[/tex]U_gravity

W_nc= 1/2mv_f+ 1/2kx²+mgh

my question is... where does friction come in... f=uN=u*mg from the equation above... and which x do I use for [tex]\Delta[/tex]U_spring... is it .35m? the compression of the spring?

 

[alphysicist]

Hi gap0063,

Homework Statement

A spring with a spring-constant 2.5 N/cm is compressed 35 cm and released. The 4 kg mass skids down the frictional incline of height 37 cm and inclined at a 20◦ angle. The acceleration of gravity is 9.8 m/s² . The path is frictionless except for a distance of 0.5 m along the incline which has a coefficient of friction of 0.3.
What is the final velocity vf of the mass?
Answer in units of m/s.

k=2.5 N/cm= 250 N/m
d=35 cm=.35m
m= 4 kg
h=37cm=.37,
theta=20◦
distance of friction = .5 m
Coefficient of friction=.3

Homework Equations

W_nc=[tex]\Delta[/tex]K+[tex]\Delta[/tex]U_spring+[tex]\Delta[/tex]U_gravity

The Attempt at a Solution

v_i=0
so W_nc=K_f+ [tex]\Delta[/tex]U_spring+[tex]\Delta[/tex]U_gravity

W_nc= 1/2mv_f+ 1/2kx²+mgh

Your kinetic energy should be 1/2 m v_f².

Also, remember that we are tracking the energy changes here. So [tex]\Delta U_{\rm spring}=\frac{1}{2}k x_f^2-\frac{1}{2}k x_i^2[/tex]. Then, x_f is the compression or stretch of the spring when the block is at the final point, and x_i is the compression or stretch of the spring when the block is at the initial point.

my question is... where does friction come in... f=uN=u*mg

The frictional force is f=uN, but the normal force is not equal to mg. Do you see what it needs to be?

from the equation above... and which x do I use for [tex]\Delta[/tex]U_spring... is it .35m? the compression of the spring?