Friction/spring/inclined plane problem

AI Thread Summary
The problem involves a 1kg block on a frictionless 5kg block, which is on a 45° inclined plane with friction, connected by a spring. A force of 200N is applied to the 5kg block, and the goal is to find the spring length at equilibrium. Initial attempts included incorrect equations and miscalculations of acceleration, particularly regarding the inclusion of spring force in the overall system. After correcting the equations and considering the vertical and horizontal accelerations, the final elongation of the spring was determined to be 7cm. The discussion highlights the importance of accurately setting up equations and considering the system dynamics.
Alex.malh
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Homework Statement


There is a block of 1kg (m1) resting frictionless on another block of 5kg (m2). m1 is connected by a horizontal spring to m2. m2 is resting on an inclined plane of 45°. Between m2 and the plane there is friction.
A force of 200N is applied on m2, pushing it upwards.

F=200N, m1=1kg, m2=5kg, f=0.2, k=100N/m, initial spring length= 0.20m, g=10m/s²

What is the spring length at equilibrium?

Homework Equations


F=m*a
F=k*spring length
Friction W= (reaction force R)*f

The Attempt at a Solution


Increase in spring length = l
I start with 4 equations:
F - w*cos45 - R*cos45 - k*l = (m1+m2)*a
m1*g + m2*g - w*sin45 - R*sin45 = 0
w=0.2*R
k*l=m1*a

Solving this i get a=20m/s² which can't be right.
 

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Why do you add k*l in 1st equation ??

You should be writing 1st equation considering both blocks together as one system . You will then not get k*l in the equation .

The rest seems fine .
 
Alex.malh said:
F - w*cos45 - R*cos45 - k*l = (m1+m2)*a
m1*g + m2*g - w*sin45 - R*sin45 = 0
As qwertywerty notes, that kl term should not be there. That is internal to the two-mass system.
But the second equation looks wrong too. Check the signs, and consider what the vertical acceleration is.
 
You're right - my bad . So you would have to consider some net acceleration and write horizontal and vertical accelarations in it's terms .
 
haruspex said:
As qwertywerty notes, that kl term should not be there. That is internal to the two-mass system.
But the second equation looks wrong too. Check the signs, and consider what the vertical acceleration is.
Thanks for the quick reply.
New equations:
F - w*cos45 - R*cos45 = (m1+m2)*ah
(m1+m2)*g + w*sin45 - R*sin45 = (m1+m2) *av
w=0.2*R

Now to solve this I'm still missing a relationship between ah and av.
I've taken av = ah ( angle is 45°, if the block has moved 1m to the right, it will have moved the same distance vertically, xv = xh -> av = ah)
Now i end up with R=495N and a=36.7m/s²

Not possible imo.
Do you guys see where i go wrong?

thanks!

regards,
 
Would you check your calculations ?
 
In your second equation, you have the wrong sign on the ma term. The blocks are accelerating vertically upward. The way you have it, av = - ah

Chet
 
Chestermiller said:
In your second equation, you have the wrong sign on the ma term. The blocks are accelerating vertically upward. The way you have it, av = - ah

Chet

Ah yes of course. Well, I've got it now. 7cm spring elongation :)
Thx a lot!
 
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