Homework Help: Mechanics problem

Work with the forces instead of energy.

ehild

 

A has to be just a bit greater than zero, so practically zero.


ehild

 

What does "shift" mean? You need to made the second block moving. The force F has to be enough to move the first block and stretch the spring. How the first block will move and what is the maximum stretching you can have with the applied constant force F? That stretching of the spring must be enough to overcome the friction of the second block.

ehild

 

Newton's second law for the first mass?

ehild

 

I reconsidered your attempt with work and energy, it might be a better approach as mine one.

Should the block of mass m have non-zero velocity when the bigger block starts moving?

But it would be useful to perform an experiment. What happens when you exert some force on mass m, bigger than f1=μmg?

ehild

 

Does anything change in the problem if the spring is replaced with a rigid link?

 

Yes, it will.

ehild

 

Does the mass m need to move at the instant when kx=uMg?

ehild

 

I'll rephrase my question: does anything change with regard to the magnitude of the force if the link is rigid?

 

Yes. The smallest force which is enough just to move the bigger mass out of its steady state is less than the force needed to move both blocks together with the same velocity (and acceleration).

See picture. What happens when you connect the weight?

ehild

 

I do not see how. To move the bigger mass, the spring has to pull it with some force F. To make the spring do it, it must be extended till its reaction force is equal to F. How is that different from a rigid link?

 

In case of a rigid link both blocks have to move together. In case of the spring the block pulled by F will start to move if F overcomes friction, but the second block stays in rest untill the displacement x of the first block yields enough force to overcome the friction of the second block. The question is what constant force is needed to displace the first block so far.

ehild

 

I Think this is the solution....
for the body to accelerate horizontally then the forces on m are
External Force F
Frictional force -mgf
Force due to the spring -kx
Force Due Friction on M Mgf


Mgf=2mgf
F-(mgf+kx)>2mgf
F>3mgf+kx
F>12+kx If k is negligible then
F>12 Newtons

 

F is the applied force. That force will do work if the block moves (that is, when F overcomes friction F>f).
If the displacement is x, the work of the applied force is Fx and the work of friction is fx.

ehild

 

F is the applied force, exerted on the first block. On the second block, the elastic force of the spring acts which is not equal to F. It depends on the displacement of the first block.

ehild

 

Nobody tried an experiment????
SO: start pulling the first block. Supposed that the force is big enough, the first block starts to accelerate. ma=F-μmg-kx. (You have seen similar equation when a weight is hanged on a vertical spring. What kind of motion is it?) The block accelerates to the right, speeds up, its displacement increases, but the spring force increases, too, so the acceleration decreases, turns to negative and the speed decreases till the block stops at maximum displacement.
The force succeeds to shift the second block when the elastic force at this maximum stretching of the spring is greater than μMg.

ehild