Minimum work done to push the cube over

[hello478]

Homework Statement

image below

Relevant Equations

w=f*d

my attempt:
centre of mass of standing block is at 0.6
and the fallen block is at 0.25
so then change in height = 0.35
and mg = 4000 N
so change in gpe = 4000* 0.35 = 1400J

but wouldnt we take into account that we only have to apply force till the centre of mass is outside of the block so would it still require the force which is equal to its weight
and what about the distance it falls forward? wouldnt that also take force?
correct answer is 200N

 

[Steve4Physics]

You are asked for the minimum energy which has to be supplied to make the block roll over.

Imagine you are doing it. You have to supply some energy to rotate the block through some angle to reach a critical position – and then the block will fall over naturally, without you having to supply further energy.

So the question is: what is the minimum energy you need to supply to get the block to the critical position?

And the underlying question is: what is the ‘critical position’?

 

[haruspex]

so change in gpe = 4000* 0.35 = 1400J

As an increase or a decrease?

but wouldnt we take into account that we only have to apply force till the centre of mass is outside of the block

How can the mass centre ever be outside the block? What did you mean?

and what about the distance it falls forward?

You are not given a horizontal distance of movement.

 

[kuruman]

How can the mass centre ever be outside the block? What did you mean?

I took that to mean that if you drop a vertical line down from the CM, it crosses the side of the inclined block and not the base.

 

[hello478]

I took that to mean that if you drop a vertical line down from the CM, it crosses the side of the inclined block and not the base.

yeah this is what i meant

 

[hello478]

You are not given a horizontal distance of movement.

so like, shouldnt it be considered?

 

[Frabjous]

Try plotting the position of the center of mass as you rotate through 90 degrees.

 

[hello478]

And the underlying question is: what is the ‘critical position’?

the critical position is option c, a little before that, as soon as cg is out of base...

in this green pic, would the block return back or fall over??

 

[haruspex]

so like, shouldnt it be considered?

It takes no more work. Once it is falling forward, job done.

 

[Frabjous]

the critical position is option c, a little before that, as soon as cg is out of base...
View attachment 342566

What is the position of the CG for these three pictures?

 

[hello478]

What is the position of the CG for these three pictures?

a) equilibrium
b) inside base
c) outside base

the green pic, tip of base
another question for green pic
would the block return back or fall over??

 

[haruspex]

the critical position is option c, a little before that, as soon as cg is out of base...
View attachment 342566

As you have drawn it, that is rather beyond the critical position. When only a tiny bit beyond the critical position, how much work have you done? Do you need to do any more?

 

[Frabjous]

the green pic, tip of base
another question for green pic
would the block return back or fall over??

Assuming zero velocity, the green pick is an unstable point like being at the top of a hill.

 

[Frabjous]

Try plotting on a grid. You need to quantify the position of the CG.

 

[Frabjous]

Another hint: the problem with your calculation is that the end state has lower potential energy than the initial state, so you are calculating energy “released”.