Final velocity of spring down incline with fricition

AI Thread Summary
The discussion focuses on calculating the final velocity of a mass sliding down an incline after being released from a compressed spring, incorporating the effects of friction. Key variables include a spring constant of 2.5 N/cm, a mass of 4 kg, and a frictional distance of 0.5 m with a coefficient of friction of 0.3. Participants emphasize the importance of tracking energy changes, specifically how to apply the work-energy principle and the correct values for potential and kinetic energy. The role of friction is clarified, noting that the normal force differs from the weight of the mass due to the incline's angle. Understanding these concepts is crucial for accurately determining the final velocity of the mass.
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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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