Calculating Retarding Force for a Free Falling Mass

[gary32]

Homework Statement

I need to work out the following;
A spherical mass of 5kg is allowed to free fall through a distance of 6m onto a shock absorbent surface without rebound. If the mass penetrates into the surface a distance of 0.6mm before stopping, find the minimum retarding focre on the mass. Use any preferred method and assume negligible air resistance during free fall.

Homework Equations

F = Mass x Acceleration ?
Acceleration = V-U
It's the equation I'm looking for that's the problem,

The Attempt at a Solution

Would it just be the gravitational constant i need for acceleration?
F = 5kg x 9.81 x 6m = 294.3
and then work out the opposite for minimum retarding force on slow down (not sure what to do now)?

Any help would be great,
Thanks

 

[serllus reuel]

Use any preferred method

I believe they are referring to momentum or energy. You seem to be onto the latter. Recall the work energy theorem.

F = 5kg x 9.81 x 6m = 294.3

That is actually work, not force. F=ma, W= ∫F dx = Fd for a constant force = mad

 

[gary32]

I believe they are referring to momentum or energy. You seem to be onto the latter. Recall the work energy theorem.

That is actually work, not force. F=ma, W= ∫F dx = Fd for a constant force = mad

Thanks for your quick responce,
So where do I go from here :/

 

[serllus reuel]

You should perhaps read the section on work and energy in your textbook.
The ball goes from rest at 6m height, to motion, to rest on the floor.
work is force times the distance which the force acts. The amount of work done is equal to the chance in energy. In this case, since air resistance is negligible, the work done by gravity is equal to the increase in kinetic energy.

So in the process of falling, the ball gained a certain amount of KE from work by gravity. As it comes to rest, all this KE must go away (and be converted into heat, sound, etc) This is the result of work done by the floor. That is why the ball cannot stop instantly; it must be in contact with the floor for some vertical DISTANCE.

In summary, conservation of energy is your friend. There is no initial or final KE (it starts from rest and ends at rest), so work in = work out, F of gravity * dist = stopping force * dist

 

[gary32]

You should perhaps read the section on work and energy in your textbook.
The ball goes from rest at 6m height, to motion, to rest on the floor.
work is force times the distance which the force acts. The amount of work done is equal to the chance in energy. In this case, since air resistance is negligible, the work done by gravity is equal to the increase in kinetic energy.

So in the process of falling, the ball gained a certain amount of KE from work by gravity. As it comes to rest, all this KE must go away (and be converted into heat, sound, etc) This is the result of work done by the floor. That is why the ball cannot stop instantly; it must be in contact with the floor for some vertical DISTANCE.

In summary, conservation of energy is your friend. There is no initial or final KE (it starts from rest and ends at rest), so work in = work out, F of gravity * dist = stopping force * dist

Thanks, I have reviewed KE,

I have now came to this,

KE = Mass x Dist x Grav = 294.3J

So, the work done on its downfall is 294.3J so that energy must be released in 0.6mm as stated so

Minimum retarding force = 294.3 / (0.6 x 1000) <-- to convert to metres, which equals
= 490500Newtons

That sound fairly reasonable answer but seems too easy :/

 

[CAF123]

Thanks, I have reviewed KE,

I have now came to this,

KE = Mass x Dist x Grav = 294.3J

So, the work done on its downfall is 294.3J so that energy must be released in 0.6mm as stated so

Minimum retarding force = 294.3 / (0.6 x 1000) <-- to convert to metres, which equals
= 490500Newtons

That sound fairly reasonable answer but seems too easy :/

It is correct (taking g = 9.81m/s²). Here is another solution:

The speed of the ball after its descent through 6m is $v = \sqrt{2gh}$. The ball now travels a distance of .6mm penetration (and is ultimately at rest) so it undergoes an acceleration $a = -u^2/2s$ where s = .6mm and $u = \sqrt{2gh}$. Hence the force applied is $F = ma = -mu^2/2s$.

 

[serllus reuel]

That seems right. The force is huge because it acts on such a short distance.

 

[serllus reuel]

@CAF123 that is an interesting approach using kinematic equations (although it is equivalent to energy)

 

[gary32]

It is correct (taking g = 9.81m/s²). Here is another solution:

The speed of the ball after its descent through 6m is $v = \sqrt{2gh}$. The ball now travels a distance of .6mm penetration (and is ultimately at rest) so it undergoes an acceleration $a = -u^2/2s$ where s = .6mm and $u = \sqrt{2gh}$. Hence the force applied is $F = ma = -mu^2/2s$.

This seems a little advanced for me however makes a little bit of sense, it did say use any preferred method so basically hinting there is more than one way as you just described,
Thanks

That seems right. The force is huge because it acts on such a short distance.

Great, I can see it all so much clearer now, I don't see why I had a problem in the first place, I think it because I am not confident with all the different equations and how 1 parameter affects another and when it should be implemented to the workings out

Thanks!

 

[serllus reuel]

This seems a little advanced for me however makes a little bit of sense, it did say use any preferred method so basically hinting there is more than one way as you just described

I was thinking of momentum, which is WAY more complicated. The above method is perhaps more fundamental as it is essentially deriving the work energy theorem for a constant force.

recall v² = v₀² +2aΔx

 

[gary32]

I was thinking of momentum, which is WAY more complicated. The above method is perhaps more fundamental as it is essentially deriving the work energy theorem for a constant force.

recall v² = v₀² +2aΔx

That looks like the formula I used in my other question

v^2 = u^2 + 2a x d

https://www.physicsforums.com/showthread.php?t=698522
Help would be appreciated if you could,

 

[haruspex]

Minimum retarding force = 294.3 / (0.6 x 1000) <-- to convert to metres, which equals
= 490500Newtons

I guess you meant 294.3 / (0.6 / 1000)

 

[gary32]

I guess you meant 294.3 / (0.6 / 1000)

Indeed I did Does everything else look okay?
Thanks :)

 

[haruspex]

Indeed I did Does everything else look okay?
Thanks :)

Yes. Mind you, I don't like the way the question is worded. What does 'minimum retarding force' mean? In general, the force will not be constant through the impact. You have calculated what the force would be if it were constant, so at different points during the impact it could be both more and less than that. I believe the question should have asked for the minimum value of the maximum retarding force.

 

[gary32]

Yes. Mind you, I don't like the way the question is worded. What does 'minimum retarding force' mean? In general, the force will not be constant through the impact. You have calculated what the force would be if it were constant, so at different points during the impact it could be both more and less than that. I believe the question should have asked for the minimum value of the maximum retarding force.

Yeah, I see what you are saying, the amount of force during different depths of penetration of the mass is changing.

 

[vadevalor]

sounds like an oxymoron lol.