Retarding impact force when object impacts stationary object

[GeorgeMJ]

Homework Statement

in a metal stamping machine, the upper die block has a mass of 35kg and falls from a height of 2m on to a fixed metal block.
if the depth of indentation is found to be 10mm, find the average stamping force assuming the upper die block does not rebound.

force=work/distance ?
power= work/time ?

hello i am very stuck on this question

work=35 x 9.8 x 2
=686J

Force= 343N

2gh= v^2

2 x 9.8 x 2 = v
v = 39.2 m/s ?

power = 17.5 w

 

[TomHart]

Welcome to Physics Forums. So it looks like you calculated the potential energy to be 686 J. That looks correct. Then you calculated Force = 343 N. That looks right too, which is the weight = (35 kg)(9.8 m/s^2) = 343 N. Then you used the potential energy to calculate v². That looks right too. That is the velocity squared of the block just before it hits the surface. So if it was me, I would probably work this using kinematic equations. You know the initial velocity, you know the final velocity, and you know the distance. So you can get the average acceleration from that. From that you could get the average force.

Or else, you could also use the force = work/distance equation that you listed. You know the distance to be 10 mm for which the average force is applied, right? And you know the amount of work that will be done to stop the block because you already calculated it. Does that make sense?

 

[haruspex]

Homework Statement

in a metal stamping machine, the upper die block has a mass of 35kg and falls from a height of 2m on to a fixed metal block.
if the depth of indentation is found to be 10mm, find the average stamping force assuming the upper die block does not rebound.

force=work/distance ?
power= work/time ?

hello i am very stuck on this question

work=35 x 9.8 x 2
=686J

Force= 343N

2gh= v^2

2 x 9.8 x 2 = v
v = 39.2 m/s ?

power = 17.5 w

Unfortunately the question is flawed. There is not enough information to work out the average force. That is because the average force is defined as $F_{avg}=\frac {\Delta p}{\Delta t}=\frac{\int F.dt}{\int.dt}$, not $\frac{\Delta E}{\Delta s}=\frac{\int F.ds}{\int .ds}$.
The two will be the same if the force is constant, but if the force is known to be constant why ask for average force?
See part 3 of https://www.physicsforums.com/insights/frequently-made-errors-mechanics-forces/.

 

[scottdave]

So actually it is 686 Joules of kinetic energy, just before it hits the metal. The metal does work on the die, slowing it down. They tell you the distance it travels while slowing. How can you use this and the Work formula to find the average force exerted to slow it down?

 

[haruspex]

How can you use this and the Work formula to find the average force exerted to slow it down?

You can't. See post #3.

 

[scottdave]

Your post is interesting @haruspex . Perhaps since the question was posted in the introductory Physics section, then the teacher who assigned was looking for something a little simpler. The kinetic energy of the moving die will be converted into work (stamping the metal) plus some heat (most likely).

 

[haruspex]

The kinetic energy of the moving die will be converted into work (stamping the metal) plus some heat (most likely).

Well, it all ends up as heat, but that is irrelevant. The point is that if the force is not constant then work/distance does not yield the average force.

 

[PeterO]

Well, it all ends up as heat, but that is irrelevant. The point is that if the force is not constant then work/distance does not yield the average force.

At introductory level, I believe that "average force" means "we know the force will vary, but what constant force would achieve the same outcome".

 

[haruspex]

At introductory level, I believe that "average force" means "we know the force will vary, but what constant force would achieve the same outcome".

Average force is well-defined. Giving the wrong impression that it can be calculated as ΔE/Δs is misleading, and leads the next generation of educators to propagate the error. It is quite easy to write instead "find the force, assuming it is constant".

 

[jbriggs444]

Average force is well-defined. Giving the wrong impression that it can be calculated as ΔE/Δs is misleading, and leads the next generation of educators to propagate the error. It is quite easy to write instead "find the force, assuming it is constant".

I bristle a bit at the idea that "average force" necessarily denotes a time-weighted average. A distance-weighted average is still an "average".

Edit: Which average you want depends on what you are going to do with the average. If you are going to multiply it by total path length to get work done then you want a distance-weighted average (and you want a scalar). If you are going to multiply it by elapsed time to get momentum transferred then you want a time-weighted average (and you want a vector).

 

[haruspex]

I bristle a bit at the idea that "average force" necessarily denotes a time-weighted average.

It's the only way to make it consistent with average acceleration. Indeed, you could have a distance-weighted average acceleration, but I think you would agree that an unqualified reference to "average acceleration" definitely means a time-weighted one.

I have checked many authoritative-looking references on the net. I only found two that appeared to accept distance-weighted. I wrote to the author of one (Prof Rod Nave, hyperphysics). He agreed it was wrong and rewrote the page.

It's an interesting exercise to compare the two different "average" forces for SHM.

I suspect that average force is not a useful concept anyway in the real world, just an easy question to devise for students. Maximum force is of greater practical interest.