Final velocity of spring down incline with fricition

In summary: Yes, it should be the compression of the spring at the final point, which is when the block is at the bottom of the incline. So it would be the total distance compressed, which is 35 cm plus the distance traveled down the incline, which is 0.5 m. Therefore, the total compression would be 0.85 m.
  • #1
gap0063
65
0

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=[tex]\Delta[/tex]K+[tex]\Delta[/tex]Uspring+[tex]\Delta[/tex]Ugravity


The Attempt at a Solution



vi=0
so Wnc=Kf+ [tex]\Delta[/tex]Uspring+[tex]\Delta[/tex]Ugravity

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 [tex]\Delta[/tex]Uspring... is it .35m? the compression of the spring?
 
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  • #2
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=[tex]\Delta[/tex]K+[tex]\Delta[/tex]Uspring+[tex]\Delta[/tex]Ugravity


The Attempt at a Solution



vi=0
so Wnc=Kf+ [tex]\Delta[/tex]Uspring+[tex]\Delta[/tex]Ugravity

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 [tex]\Delta U_{\rm spring}=\frac{1}{2}k x_f^2-\frac{1}{2}k x_i^2[/tex]. 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 [tex]\Delta[/tex]Uspring... is it .35m? the compression of the spring?
 

Related to Final velocity of spring down incline with fricition

1. What is the equation for calculating the final velocity of a spring down an incline with friction?

The equation for calculating the final velocity of a spring down an incline with friction is Vf = √(2kx/m (sinθ - μcosθ)) where Vf is the final velocity, k is the spring constant, x is the displacement of the spring, m is the mass of the object attached to the spring, θ is the angle of the incline, and μ is the coefficient of friction.

2. How does the angle of the incline affect the final velocity of the spring?

The angle of the incline affects the final velocity of the spring by changing the amount of gravitational potential energy that is converted into kinetic energy. A steeper incline will result in a higher final velocity, while a shallower incline will result in a lower final velocity.

3. What role does friction play in the final velocity of the spring down an incline?

Friction plays a significant role in the final velocity of the spring down an incline. It opposes the motion of the object and reduces the amount of kinetic energy that is converted from the potential energy of the spring. This results in a lower final velocity compared to a scenario with no friction.

4. Can the final velocity of the spring be greater than the initial velocity?

Yes, in some cases the final velocity of the spring can be greater than the initial velocity. This can happen when the angle of the incline is steep enough and the coefficient of friction is low enough to allow for a greater conversion of potential energy into kinetic energy.

5. Are there any assumptions made in the calculation of the final velocity of the spring down an incline with friction?

Yes, there are a few assumptions made in the calculation of the final velocity of the spring down an incline with friction. These include assuming that the incline is smooth and that the spring is ideal, meaning it obeys Hooke's Law and does not experience any energy losses due to heat or other factors. Additionally, the calculation assumes that the object attached to the spring does not experience any external forces or resistances during its motion down the incline.

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