Block moving down an inclined plane, spring problem.

In summary, the problem involves finding the amount of compression in a spring after a block with given values (mass, initial velocity, kinetic coefficient of friction) comes to rest. The solution involves calculating the velocity at which the block touches the spring, using a free body diagram to find the acceleration, and then using an energy balance equation to solve for the compression length of the spring. The correct answer is 0.2798m and the acceleration should be less than gravity.
  • #1
Dealucis
3
0

Homework Statement



problem1.jpg


Find the amount which the spring is compressed after the mass comes to rest. Assume a kinetic coefficient of friction of 0.11 and an initial velocity along the surface of v = 4m/s.

Given values :
mass = 2.2kg
k = 500 N/m
delta = 0.036m
theta = 23 degrees

The attempt at a solution

Alright.. here goes

My approach to this problem is to find the velocity at where the block touches the spring, and make it as the initial velocity.
------------------------------------------------------------------------------------------
Using a free body diagram, I got the following values.

F = mg = 21.56N
N = mg cos theta = (21.56)cos23 = 19.846N
friction force = N.coefficient = (19.846)(0.11) = 2.183N
force down the plane = mg sin theta = (21.56)sin23 = 8.424N

Adding up the force in the x direction (parallel to the slope), F = 8.424 + (-2.183) = 6.241N

since F = ma,
6.241N = (2.2kg)a
therefore, a = 2.837m/s^2 [sorry, I'm not familiar with the forums. the unit is metre per seconds squared]
------------------------------------------------------------------------------------------
Now, with the acceleration, the velocity at when it touches the spring can be calculated :

final velocity squared = initial velocity squared + 2(acceleration)(distance)
v^2 = u^2 + 2(a)(r)
v^2 = 4^2 + 2(2.837)(0.036)
v = squareroot(16 + 0.204)
v = 4.025m/s

therefore, the velocity at when it touches the spring is 4.025m/s
------------------------------------------------------------------------------------------
Using the work-energy equation,

T1 + U = T2

as when the final velocity is when the block comes to rest, the final velocity is 0, therefore

(mv^2)/2 + U = 0
17.821 +U = 0
therefore U = 17.821J
------------------------------------------------------------------------------------------
Using the value, the compression length can be found:
17.821 = 0.5(k)(final compression length squared - initial compression length squared)
17.821 = 0.5(500)(x^2 - 0)
17.821 = 250(x^2)
0.071283 = x^2
x = squareroot(0.071283)
x = 0.267m

that was my answer, but it's wrong. Much help would be appreciated on where i went wrong and how to solve the problem. Thanks.
 
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  • #2
Using the work-energy equation,

T1 + U = T2

as when the final velocity is when the block comes to rest, the final velocity is 0, therefore

(mv^2)/2 + U = 0
17.821 +U = 0
therefore U = 17.821J

What is T2 and why is it zero? Where are you considering the potential energy of the block in this energy balance?

If I was to go about this problem, I would do it in two steps. First would be to find the velocity of the block at the spring. a = (Fg - Ff)/m

Then I would do an energy balance of the spring and block as a system. KEblock + PEblock = PEspring

Which you would solve for the displacement in the PEspring term.
 
  • #3
Topher925 said:
What is T2 and why is it zero? Where are you considering the potential energy of the block in this energy balance?

If I was to go about this problem, I would do it in two steps. First would be to find the velocity of the block at the spring. a = (Fg - Ff)/m

Then I would do an energy balance of the spring and block as a system. KEblock + PEblock = PEspring

Which you would solve for the displacement in the PEspring term.

Thanks for the tip, but i don't quite get your formula for acceleration. Can you explain more?

For the potential energy, yeah, i forgot about that. After revising my calculations, i still got the wrong answer.

Kinetic energy of block = 0.5mv^2 -0.5mu^2 (where v = final velocity and u = initial velocity)
therefore kinetic energy = -17.821
potential energy of block = -mg(0.036sin23) = -0.303

potential energy of spring = -0.5kx^2

solving the equation I ended up with 0.269m, which is still wrong :(
 
  • #4
The solution is 0.2798m.

How is it possible? working backwards I found the acceleration to be around 21m/s^2. Isn't the acceleration suppose to be less than gravity? Can someone help me point out where i went wrong please? (please refer to my calculations above for my steps on finding the acceleration). Much help would be appreciated!
 
  • #5
Hi Dealucis,

Dealucis said:
Thanks for the tip, but i don't quite get your formula for acceleration. Can you explain more?

For the potential energy, yeah, i forgot about that. After revising my calculations, i still got the wrong answer.

Kinetic energy of block = 0.5mv^2 -0.5mu^2 (where v = final velocity and u = initial velocity)
therefore kinetic energy = -17.821
potential energy of block = -mg(0.036sin23) = -0.303

I don't believe this is correct. You're writing the energy equation for the motion while the spring is being compressed; but the distance 0.036 refers to the part of the motion that came before that. What do you think this term should be?

potential energy of spring = -0.5kx^2

solving the equation I ended up with 0.269m, which is still wrong :(

Dealucis said:
The solution is 0.2798m.

How is it possible? working backwards I found the acceleration to be around 21m/s^2. Isn't the acceleration suppose to be less than gravity? Can someone help me point out where i went wrong please? (please refer to my calculations above for my steps on finding the acceleration). Much help would be appreciated!

I'm not seeing where you are getting the acceleration to be 21m/s^2 if you use that answer. Can you give more details?
 
  • #6
------------------------------------------------------------------------------------------
Using the work-energy equation,

T1 + U = T2

as when the final velocity is when the block comes to rest, the final velocity is 0, therefore

(mv^2)/2 + U = 0
17.821 +U = 0
therefore U = 17.821J
------------------------------------------------------------------------------------------

You are missing the loss of energy in this part.

So, the equation should be U-T1=Wf (where U = potential energy of spring, T1=total initial energy of object, Wf= work done by friction)

U is equal to T2 ( T2 = total final energy of object)
It is because when the spring is compress by object, so the spring will gain the potential energy from the total final energy of object.

0.5*2.2*(4.025^2) + 2.2*9.8*(x*sin(23))=2.185*x + 0.5*500*x^2

From above equation u will get the value of x which is the amount which the spring is compressed.

x= 0.2798m
Here is the answer.
 

1. What is the concept of a block moving down an inclined plane, spring problem?

The block moving down an inclined plane, spring problem is a classic physics problem that involves a block sliding down an inclined plane with a spring attached to it. The goal is to determine the acceleration of the block and the compression of the spring at equilibrium.

2. How do you calculate the acceleration of the block in this problem?

The acceleration of the block can be calculated using Newton's second law of motion, which states that the net force on an object is equal to the mass of the object multiplied by its acceleration. In this problem, the net force acting on the block is the component of the force due to gravity parallel to the inclined plane, which can be calculated using trigonometry.

3. What role does the spring play in this problem?

The spring in this problem acts as a restoring force, meaning it exerts a force on the block in the opposite direction of its displacement. This force is directly proportional to the amount of compression or extension of the spring, as described by Hooke's law.

4. How does the angle of the inclined plane affect the acceleration of the block?

The angle of the inclined plane affects the acceleration of the block because it determines the component of the force of gravity acting on the block parallel to the plane. As the angle increases, the component of the force of gravity also increases, resulting in a greater acceleration of the block down the plane.

5. What factors can change the equilibrium position of the block and the spring in this problem?

The equilibrium position of the block and spring can be changed by altering the mass of the block, the spring constant of the spring, or the angle of the inclined plane. Additionally, external forces such as friction can also affect the equilibrium position of the block and spring.

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