A 15 LB weight is dropped vertically 1 ft

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Homework Help Overview

The discussion revolves around a scenario where a 15 LB weight is dropped vertically from a height of 1 foot and is stopped instantaneously. Participants are exploring the implications of this situation on the force experienced at impact, particularly in relation to inertia and momentum.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of force at impact, referencing momentum and energy concepts. Some suggest using equations related to velocity and kinetic energy, while others question the feasibility of instantaneous stopping and its implications for force.

Discussion Status

The discussion is ongoing, with various interpretations being explored. Some participants have provided guidance on using momentum and energy principles, while others highlight the complications introduced by the concept of instantaneous stopping, suggesting it leads to an infinite force scenario.

Contextual Notes

There are assumptions about the nature of the stopping process and its physical implications, particularly regarding the definition of force in relation to time. The discussion reflects a mix of intuitive reasoning and formal physics concepts.

Nornh
A 15 LB weight is dropped vertically 1 ft. and is stopped instantaneously. What is the weight at impact due to inertia?

That wasn't quite the answer I needed.


Question:

From "0" velocity the 15lb object is released vertically at a 1 ft drop and then it is instantaneously stopped. What is the force in lbs of the object at that point?
 
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A 15 LB weight is dropped vertically 1 ft. and is stopped instantaneously. What is the weight at impact due to inertia?

hey,
so far i understand , you need to find the change in momentum per unit time that will give you the force i mean the weight at impact.

the initial momentum is 0 becasue initially the weight was at rest.

for the final momentum P=mv you need v to calculate.

so use the equation v^2= v0^2 + 2gh
v0= the initial velocity =0, g acceleration due to gravity and h the distance it travelled.
i hope this will help you.

bye
SKT
 
Another way to find the speed after the weight has fallen 1 foot is to use kinetic energy. When the weight is first released, it has 0 kinetic energy and potential energy 15 pounds*1 foot= 15 foot-pounds.

After it has fallen the 1 foot, it's potential energy is 0 and it's kinetic energy is (1/2)mv2= 15. The mass of a 15 pound weight is 15/32 poundas so (15/64)v2= 15. v2= 64 and v= 8 ft/sec.

In stopping, it changes speed by -8 ft/sec (from 8 to 0).

HOWEVER! You have a serious problem- Force is "change in speed over change in time" and there is NO "change in time". You have specified that the weight is stopped "instantaneously". That can be done only by an infinite force!
 
hi HallsofIvy


thanks, so what do you suggest then,

what should be the weight at the time of imapct? is it infinite ?
SKT
 
I feel there is some pre-scientific thinking going on here. Children observe that a rock makes a bigger crater in the sand if it is dropped from a greater height. So they believe that a moving body is heavier than a resting body. I think this was a widely accepted concept in ancient times, too.
Newtonian mechanics claim that the larger crater is due to the larger inertial force required to stop the rock, and not due to an increased mass of the rock.
You could ask, what is the force exerted by the falling body onto the ground. But if the stopping is 'instantaneous', the answer would be an 'infinite' force via Newton's law F = ma.
 

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