If a 63 pound object is dropped from five feet on to a hand, how much force was exerted on the hand?

Question

Please help if you can. I need pounds of force on the hand.

Answer 1

Brian wrote:
> .....it doesnt matter how high you drop it from.
Brian, Brian, Brian. If the force is the same regardless of how high you drop it from, then I suppose you won't mind if I drop it on your hand from 50 feet instead of 5?

Brian calculated the force that the earth exerts on the object. This matches the force that the object exerts on your hand ONLY in the case where the object is not accelerating. But in this case the object IS accelerating (actually DEcelerating) as your poor hand causes it to slow from 18 ft/sec down to zero.

paulred2 wrote:
> acceleration=distance x force of gravity(9.8)
That's a completely erroneous formula.

princess wrote:
> 153.2*1.5=229.8 newtons (weight after the fall)
No! You can't multiply the weight times the distance to get the "weight after the fall." There is no such thing. This formula is baloney.

To get the REAL answer, you have to determine the rate of DECELERATION of the object after it hits your hand, and multiply that by the object's mass, according to the formula F=ma

The object is traveling at about 18 ft/sec when it hits your hand (this is sqrt(2g*5ft)).

It presumably decelerates to zero ft/sec. To figure the rate of deceleration, we must either know the amount of time it took to slow down, or the distance over which it slowed down. We're not given either one of those. However, we can assume that the _maximum_ distance of deceleration must be about the thickness of a hand--let's take that as 1 inch. In that case, the formula is:

acceleration = v² / (2d)
= (18ft/sec)² / (2 inches)
= 592.5 ft/sec²

This is about 60 times the acceleration of gravity, which means the force felt will be about 60 times the weight of the object (about 3,800 pounds).

If the object stops short of completely flattening the hand, then the value of "d" is reduced, and the corresponding force is _increased_. So 3,800 lbs. is actually a minimum.

Answer 2

First convert pounds to kilograms. Once you have that number multiply it by 9.8m/s^2 That is how much force will be exerted on your hand...it doesnt matter how high you drop it from.

EDIT:
RickB (look below to see what RickB wrote) has a point, but it is incorrect. He calculated the force that your hand would have to exert on the object in order to stop its motion within 1 inch of movement not the force that the object puts on your hand. Notice a "distance" requiered to make this calculation. This means that your hand will exert an energy ( energy = force x distance). The reason why you wouldnt want to catch an object drop from 50 feet is that it will exert a ton of kinetic energy to your hand...energy is the thing the destroys. The force will be the same all the time, the energy of the falling object on the other hand will increase with velocity. Nice try RickB, but you do not understand the subtle details of the Force. Im not saying that i do, but in this case i happen to be correct.

Answer 3

force=mass x acceleration
mass=63
acceleration=distance x force of gravity(9.8)=49
force=63 x 49
force=3087pounds of force overall

{also force divided by area(ie bottom of falling object)=pressure per unit sq}
but common sense tells you a 63 pound object solid is the weight of the average 12 year old and would crush your hand if falling from any height...

Answer 4

a 143 pound object exerts a 143 pound force, whether it is sitting still or dropped from 1000 feet.

Answer 5

first you convert pounds in kilogramms and the feet in meters. 63 lbs=14kg 5ft=1.5m
then you muliply by 9.8m/s^2 to find the acceleration 14kg*9.8m/s^2=153.2 (weight before the fall)
then you find the force exerced on the hand when the object touch the hand after a 5ft(1.5m) fall
153.2*1.5=229.8 newtons (weight after the fall)