Two blocks and a massless horizontal spring question?

[nitin4422]

1. Two masses of m1 and m2 is connected by a massless undeformed spring rest on a horizontal plane.Find the minimum constant force Fmin that has to be applied on m2 so that other block get shifted if D is the coefficient of friction between blocks and surface



2. Homework Equations -spring force kx



3. i have tried to use conservation of energy for this question as block m1 will have potential energy D^2mi^2g^/k and when it will start moving its potential energy will change into the kinetic energy ...please help i am really stuck on this question

 

[CWatters]

Wrong approach. Just look at the forces...

Ignore mass 2 for the moment. What force must the spring exert on mass 1 to move it?

Then draw a free body diagram for mass 2.

Edit: Sorry I had the numbers 1 and 2 the wrong way around when I first replied. Have fixed that now.

 

[nitin4422]

ok spring force required to move the mass m1 will be kx greater than or equal to Dm1g .now after drawing the free body diagram of mass 2 i have got F-Dm2g-kx=m2a (here a is the acceleration of mass 2).now if i put the value of kx from equation 2 to equation 1 i am getting the answer with m2a in it but the given answer is (m1/2+m2)Dg

please help

 

[Doc Al]

An energy approach is the right way to go, but you'll need to analyze forces as well. Hint: For m1 to start moving, what must be the compression of the spring?

 

[DEvens]

ok spring force required to move the mass m1 will be kx greater than or equal to Dm1g .now after drawing the free body diagram of mass 2 i have got F-Dm2g-kx=m2a (here a is the acceleration of mass 2).now if i put the value of kx from equation 2 to equation 1 i am getting the answer with m2a in it but the given answer is (m1/2+m2)Dg

please help

I disagree with the given answer.

To move block 1 at constant speed you need to shove it with force m1 D g.
To move block 2 ditto you need force m2 D g.
To move both blocks you need (m1 + m2) D g. And the fact they are connected by a spring won't make any difference. With both blocks at constant speed the spring is constant length.
Dan

 

[Doc Al]

I disagree with the given answer.

To move block 1 at constant speed you need to shove it with force m1 D g.
To move block 2 ditto you need force m2 D g.
To move both blocks you need (m1 + m2) D g. And the fact they are connected by a spring won't make any difference. With both blocks at constant speed the spring is constant length.
Dan

You want the minimum force that will make the second block (m1) just start to move.

 

[haruspex]

CWatters and DEvens, you're both overlooking that since the force required must be more than m₂Dg, m₂ will acquire some momentum. That will later assist in moving m₁.
nitin4422, pls post your working.

 

[Tanya Sharma]

Shouldn't the coefficient of friction between the blocks and the surfaces be different ? For the force applied on m2 ,we will be dealing with kinetic friction .While for block m1 to just move ,we have to deal with static friction .

 

[haruspex]

Shouldn't the coefficient of friction between the blocks and the surfaces be different ? For the force applied on m2 ,we will be dealing with kinetic friction .While for block m1 to just move ,we have to deal with static friction .

It would seem that in this question the two are to be taken to have the same value.

 

[nitin4422]

I took the blocks and spring as one system and calculated acceleration on each box a=F-D(m1+m2)g/(m1+m2) then i tried to solve it from a non inertial frame with a a psuedo force on mass m1 and used conservation of energy i tried this but no answer instead i found maximum elongation in the string 2m1{F-D(m1+m2)g}/m1+m2 sorry if its totally wrong

 

[Doc Al]

I took the blocks and spring as one system and calculated acceleration on each box a=F-D(m1+m2)g/(m1+m2)

Realize that the blocks do not have the same acceleration. You cannot treat the system as a single rigid body.

then i tried to solve it from a non inertial frame with a a psuedo force on mass m1 and used conservation of energy i tried this but no answer instead i found maximum elongation in the string 2m1{F-D(m1+m2)g}/m1+m2 sorry if its totally wrong

Please answer the question I asked in post #4:

Hint: For m1 to start moving, what must be the compression of the spring?

 

[nitin4422]

Realize that the blocks do not have the same acceleration. You cannot treat the system as a single rigid body.



Please answer the question I asked in post #4:

i think force F on the mass2 will stretch the spring to the right side of the block of mass1 therefore block of mass 1 will have a tension force kx due to the spring so in order to move the mass spring force will have to apply force equal to or greater to Dm1g therefore kx> or equal to Dm1g so x must be> or equal to Dm1g/k so.. i think compression should be equal to Dm1g/k am i thinking right?

 

[Doc Al]

i think force F on the mass2 will stretch the spring to the right side of the block of mass1 therefore block of mass 1 will have a tension force kx due to the spring so in order to move the mass spring force will have to apply force equal to or greater to Dm1g therefore kx> or equal to Dm1g so x must be> or equal to Dm1g/k so.. i think compression should be equal to Dm1g/k am i thinking right?

Excellent. Now realize that that compression distance is the distance that block 2 must move through and thus the distance over which the applied constant force F must act. Now it's time to apply energy methods.

 

[DEvens]

You want the minimum force that will make the second block (m1) just start to move.

The "given" answer is (m1/2 + m2) Dg, when pulling on mass m2.

Suppose that mass m1 is very much larger than m2, such that you can ignore m2 in the result. The given answer is thus claiming that, somehow, you only need to pull half as hard as you'd need to in order to move m1 without the spring.

Or, to take a really extreme case, let m2 be zero. So this is claiming that by pulling on the other end of the spring it requires only half the force required to move m1 as by pulling it directly.

What?
Dan

 

[Doc Al]

The "given" answer is (m1/2 + m2) Dg, when pulling on mass m2.

Imagine the masses are arranged, from left to right, as m2, spring, m1. Your task is to find the minimum constant force to push m2 so that m1 just begins to move. To find the minimum force, you will take advantage of any speed you give to m2 when you push it.

Suppose that mass m1 is very much larger than m2, such that you can ignore m2 in the result. The given answer is thus claiming that, somehow, you only need to pull half as hard as you'd need to in order to move m1 without the spring.

Yep. Realize that since the force of the spring is not constant, that your constant applied force ends up giving m2 speed with which to further compress the spring.

Or, to take a really extreme case, let m2 be zero. So this is claiming that by pulling on the other end of the spring it requires only half the force required to move m1 as by pulling it directly.

Let's just make m2 be very small. (Not sure it makes sense to have it zero, but a small m2 makes your point just the same.) The same considerations as above apply.

The trick is that you are applying a constant force that is greater than the spring compression force (at least much of the time). So you are building up speed that can end up compressing the spring more and moving m1.

 

[haruspex]

The "given" answer is (m1/2 + m2) Dg, when pulling on mass m2.

Suppose that mass m1 is very much larger than m2, such that you can ignore m2 in the result. The given answer is thus claiming that, somehow, you only need to pull half as hard as you'd need to in order to move m1 without the spring.

Or, to take a really extreme case, let m2 be zero. So this is claiming that by pulling on the other end of the spring it requires only half the force required to move m1 as by pulling it directly.

What?
Dan

It says pushing, but it also works for pulling.
As Doc Al says, you can be misled by setting m₂ to zero. As m₂ becomes very small, its initial acceleration (when there's no resistance from the spring yet) becomes arbitrarily large. Since the force will always be at least m₁/2, the energy invested in pulling/pushing distance x is at least m₁x/2. When x is small, that must exceed the energy going into the spring, kx²/2. So no matter how small you make m₂, there's a lower bound on the KE it gets.
Ever seen a bungee cord used?

 

[Tanya Sharma]

Hello nitin4422

Were you able to solve the problem ? If not , what difficulty are you facing ?