Friction problem -- 2 Blocks Sliding on a Surface

AI Thread Summary
The discussion focuses on a physics problem involving two blocks with different masses and a friction coefficient, where a force is applied to one block. The key questions are determining the time it takes for the lighter block to reach the left extremity of the heavier block and calculating the displacement of the heavier block. Participants suggest using free body diagrams and Newton's second law to analyze the forces and accelerations acting on each block. There is emphasis on understanding relative motion and the need to apply kinematic equations appropriately. The conversation highlights the challenges in calculating accelerations due to the accelerated frame of reference.
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Homework Statement


A block of mass m=2kg and one of mass M=8kg have initial velocity zero.
The friction coeffecient between m and M blocks is 0.3
Between block M and the surfasse there is no friction.
L=3 metres
F is a force constant and horizontal with magnitude 10 N, applied on block m and so block m starts to move and goes to left extremity of block M.
block M also moved as you can see in the image

1) How long does it takes to block m to arrive to left extremity of block M?

2? What is the value of the displacement of block M?

the image is:
http://s3.amazonaws.com/answer-board-image/79698f93-5a84-4194-ab39-782abe24edcd.png

Homework Equations

The Attempt at a Solution


[/B]
i put the forces applied in each block:
block 1:
x: ma1=F-F(of friction)
y: mg=Normal force 1

block 2:
x: Ma2=F(of friction)
y: Mg=Normal force 2

and then i took a2 (acceleration of block 2) = 0.735 m s^-2

i tried to use v^2=v^2 (initial) +2ax
but i don't know v neither x...any suggestion?
 
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rf1 said:

Homework Statement


A block of mass m=2kg and one of mass M=8kg have initial velocity zero.
The friction coeffecient between m and M blocks is 0.3
Between block M and the surfasse there is no friction.
L=3 metres
F is a force constant and horizontal with magnitude 10 N, applied on block m and so block m starts to move and goes to left extremity of block M.
block M also moved as you can see in the image

1) How long does it takes to block m to arrive to left extremity of block M?

2? What is the value of the displacement of block M?

the image is:
http://s3.amazonaws.com/answer-board-image/79698f93-5a84-4194-ab39-782abe24edcd.png

Homework Equations

The Attempt at a Solution


[/B]
i put the forces applied in each block:
block 1:
x: ma1=F-F(of friction)
y: mg=Normal force 1

block 2:
x: Ma2=F(of friction)
y: Mg=Normal force 2

and then i took a2 (acceleration of block 2) = 0.735 m s^-2

i tried to use v^2=v^2 (initial) +2ax
but i don't know v neither x...any suggestion?

What about looking first at the relative motion of block m with respect to block M?
 
Although you have an image, I suggest you to draw a free body diagram representing all the forces that are involved in the situation (in the image just appears one). Then see what happens. Is there constant velocity?

For there to be motion of something, there must be a force acting on that something.
 
rf1 said:
and then i took a2 (acceleration of block 2) = 0.735 m s^-2
Good. What's the acceleration of block 1?

rf1 said:
i tried to use v^2=v^2 (initial) +2ax
but i don't know v neither x...any suggestion?
Well, you do know the distance... the relative distance. Since you don't know v, you might want to choose another kinematic formula.
 
Doc Al said:
Good. What's the acceleration of block 1?Well, you do know the distance... the relative distance. Since you don't know v, you might want to choose another kinematic formula.

I can't calculate acceleration of block 1 because it is in an accelerated frame and so the second law of Newton is not valid

I tried to use Energy-Work Theorem but i can't because i don't have enough information to use it
 
rf1 said:
I can't calculate acceleration of block 1 because it is in an accelerated frame and so the second law of Newton is not valid

I tried to use Energy-Work Theorem but i can't because i don't have enough information to use it

Suppose two racing cars started at the same time. One accelerated at ##a \ ms^{-2}## and one accelrated at ##b \ ms^{-2}## and you measured the distance between them over time.

Suppose you repeated this and the first car failed to start and the second car accelerated at ##(b-a) \ ms^{-2}## and you measured the distance between them over time.

Would there be any difference?
 
rf1 said:
I can't calculate acceleration of block 1 because it is in an accelerated frame and so the second law of Newton is not valid
You can calculate the acceleration of block 1 exactly as you calculated the acceleration of block 2: By applying Newton's 2nd law from the inertial frame of the ground. Those accelerations will be with respect to the ground. It's up to you to figure out the relative acceleration of block 1 with respect to block 2.

rf1 said:
I tried to use Energy-Work Theorem but i can't because i don't have enough information to use it
You won't need the work-energy theorem; stick to Newton's 2nd law and kinematics.
 
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