If a metal can falls to the floor 3 feet, how much force/weight does it gain before it actually lands?

Question

What would be the actual weight it feels like when it strikes one's foot ? We are talking about a 12oz can of Campbell's Soup here

Answer 1

One way to find out what it would actually feel like is-- figure out the average force exerted by by your foot to stop the can. Say your foot compresses by 1cm as the can stops. Say the can has 0.5kg of mass (should be in that ballpark...) If it falls 1m (about three feet), it should gain 0.5m*10m/s^2*1m of energy-- 5 Joules of energy. Now, if it stops in 1cm-- your foot has to do some work on the can-- W=f*distance. So, f=W/Distance-- 5J/0.01m-- 500N. So, the can could "feel" like it weighs 500N, or 50kg (110 lbs.) when it hits your foot, because that's the force your foot has to exert to stop the can.

Answer 2

I don't think you meant to ask the question you did; it makes little sense. Here's what I think you meant:

If a metal can falls to the floor 3 feet, how much kinetic energy does it gain before it actually lands?

In 3 feet, the weight of something is essentially the same; only the most precise, highly complex lab tools would be able to measure any difference. Weight is simply W = mg; where g = a constant = 32.2 ft/sec^2 at Earth's surface. (By the way, I can show g is still 99.7% of that 32.2 at 30,000 feet altitude; so you can see why 3 feet is insignificant when it comes to weight.)

Weight is just the name we give the force due to gravity. But it is a force and force (F) acting over a distance (d) is called work = E = Fd = Wd = mgd where another name for work is energy (E). And as that weight drops a distance d = 3 feet, it creates work or energy, in this case kinetic energy (KE).

Thus, E = KE = 1/2 mv^2 = mgd = Wd; so we have v = sqrt(2gd); where v = the velocity at which the metal can will hit the floor when dropped from a distance = 3 feet. Notice that the velocity depends only on g and d, the mass of the can, and consequently its weight that depends in part on mass, have no bearing on how fast the can hits the floor (if no other forces, like drag force, are present).

On the other hand, we do need to know the can's mass to find the KE = 1/2 mv^2. By the way, just before the can is release at d, it has potential energy = PE = mgd. By the conservation of energy law, KE = PE when that can hits the floor. That is, the PE is converted into KE; thus the total energy is conserved, but converted.

Answer 3

Technically, the metal has no practical weight as it falls because it reaches near 0 G. The kenetic force is rapidly increased as it falls. Force is measured in ft/lbs or likewise, so, as you haven't given the surface of which strikes the foot, I can't give you a mathmatical answer. A smaller surface area would have greater impact, which is why a blunt impact from a can doesn't dent it, while an impact on an edge can dent it more than a little bit.

Answer 4

The weight it gains is negligible compared to the kinetic energy it gains as it falls. The force it gains is equal it the kinetic energy it gains during the fall.