Final velocity of spring down incline with fricition

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SUMMARY

The final velocity of a 4 kg mass released from a compressed spring with a spring constant of 250 N/m and sliding down a frictional incline can be calculated using the work-energy principle. The incline is 37 cm high at a 20° angle, with a frictional distance of 0.5 m and a coefficient of friction of 0.3. The relevant equations include the work done by non-conservative forces and the changes in kinetic, spring, and gravitational potential energy. The correct compression distance for the spring potential energy calculation is 0.35 m, the initial compression.

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gap0063
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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


Wnc=\DeltaK+\DeltaUspring+\DeltaUgravity


The Attempt at a Solution



vi=0
so Wnc=Kf+ \DeltaUspring+\DeltaUgravity

Wnc= 1/2mvf+ 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 \DeltaUspring... is it .35m? the compression of the spring?
 
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Hi gap0063,

gap0063 said:

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


Wnc=\DeltaK+\DeltaUspring+\DeltaUgravity


The Attempt at a Solution



vi=0
so Wnc=Kf+ \DeltaUspring+\DeltaUgravity

Wnc= 1/2mvf+ 1/2kx2+mgh

Your kinetic energy should be 1/2 m vf2.

Also, remember that we are tracking the energy changes here. So \Delta U_{\rm spring}=\frac{1}{2}k x_f^2-\frac{1}{2}k x_i^2. Then, xf is the compression or stretch of the spring when the block is at the final point, and xi 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 \DeltaUspring... is it .35m? the compression of the spring?
 

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