Block moving down an inclined plane, spring problem.

  • Thread starter Dealucis
  • Start date
  • #1
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.
 

Answers and Replies

  • #2
1,565
7
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
3
0
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
3
0
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
alphysicist
Homework Helper
2,238
2
Hi Dealucis,

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 :(

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
2
0
------------------------------------------------------------------------------------------
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.
 

Related Threads on Block moving down an inclined plane, spring problem.

Replies
7
Views
2K
  • Last Post
Replies
22
Views
6K
Replies
3
Views
963
Replies
2
Views
620
Replies
5
Views
7K
Replies
3
Views
10K
  • Last Post
Replies
9
Views
11K
Replies
2
Views
3K
Replies
1
Views
6K
  • Last Post
Replies
17
Views
2K
Top