Friction problem -- 2 Blocks Sliding on a Surface

In summary: But if you insist on using the work-energy theorem, you can use the fact that the final kinetic energy of block 1 is 0, as it is at rest. You can use that to find its initial kinetic energy, and work backward to find its initial speed. Then you'll be able to use the equations of constant acceleration to find the time needed to cover the 3m distance.
  • #1
rf1
3
0

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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  • #2
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?
 
  • #3
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.
 
  • #4
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.
 
  • #5
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
 
  • #6
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?
 
  • #7
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.
 

Related to Friction problem -- 2 Blocks Sliding on a Surface

1. What is friction and how does it affect the movement of two blocks on a surface?

Friction is the force that opposes the relative motion of two surfaces in contact. In the case of two blocks sliding on a surface, friction works to slow down or prevent the movement of the blocks.

2. What factors affect the amount of friction between two blocks?

The amount of friction between two blocks is affected by the type of surfaces in contact, the force pressing the surfaces together, and the roughness of the surfaces.

3. How can friction be reduced in the scenario of two blocks sliding on a surface?

Friction can be reduced by using a lubricant between the two surfaces or by making the surfaces smoother.

4. What is the difference between static and kinetic friction in the context of two blocks sliding on a surface?

Static friction is the force that opposes the start of motion between two surfaces, while kinetic friction is the force that opposes the ongoing motion between two surfaces.

5. How does the coefficient of friction play a role in the movement of two blocks on a surface?

The coefficient of friction is a measure of the frictional force between two surfaces. A higher coefficient of friction means there is a greater resistance to movement between the two blocks, while a lower coefficient of friction allows the blocks to move more easily.

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