Force due to a freely falling body?

Question

If you have a block of mass 100kg, initially at rest and drop it 10m and it impacts the ground and does not bounce it just lands flat. What are the forces it imparts on the ground and what are the forces the ground imparts on it.

I have tried simply v=(gt)^.5 and d=.5 gt^2 and solved for the finial velocity and time and calculated acceleration = change in v/ change in t, which I get the force of gravity F=ma then W=mg and the total F= ma-mg which turns out to be zero. What am I doing wrong?

Answer 1

♪♪♪♪ kid! v=gt and not (gt)^.5;
♠ what can we do with your data?
♣ initial potential energy W=mgd, m=100kg, g=9.8m/s^2, d=10m, while final kinetic energy E= 0.5m*v^2, v is final speed; now according to conservation law E=W, thus mgd=0.5m*v^2, hence v=(2gd)^0.5= 14m/s;
♦ time of fall t=v/g = 14/9.8 =1.43 s; now check d=0.5g*t^2 =10m is OK!
♥ while impact the block will loose its speed to zero in a short period of time τ, thus average deceleration α=v/τ; and according to second Newton’s law the force of impact f=m·α block’s weight p=m·g; and after time τ p=m·g is the only force remained. These forces are imparted on each other!
☻the lost kinetic energy would be translated into heat warming the spot if impact.
Here my imagination is out.

Answer 2

Hi... GOOD question.

It got me stumped for a minute.

Now if the object stopped instantenuously at the ground without bouncing, it would mean an infinite acceleration and an infinite force (going from 14m/s to 0 in 0 second...). So objects will either bounce or make an indentation. In your case, it will make a dent. How deep is the dent is crucial in measuring the force.

Ok let's imagine the object makes a 0.001 m dent (1 millimeter). Then it went from 14 m/s to 0 in a distance of 0.001m.

vf = final velocity = 0
a = acceleration
t = time
v = initial velocity = 14

vf= at + v
0 = at + v
t = -v/a

d= 1/2at^2 + vt
d= 1/2 a (-v/a)^2 + v(-v/a)
d= 1/2 v^2/a - v^2/a
d= -1/2 v^2/a
a = -1/2 v^2/d = -1/2 * 14^2 / 0.001 =
a = - 98000 m/s^2

So the object had a deceleration of 98000 m/s^2. The force felt by the object and the ground was 98000m/s^2 * 100kg = 9 800 000N

Check out the attached website it has a neat calculator for this.

Answer 3

You have a problem with the right choice of equations of motion. Since time is of no consequence here, use the eqn v^2 = u^2 + 2*g*h, where u = initial vel., v = end vel., g = acceleration due to gravity, h = distance travelled.
Since motion started from rest, u = 0; g = 9.81 m/sec^2
Therefore, v^2 = 2* 9.81 * 10
When the mass of weight 100 kg hits the ground with a velocity of v, it imparts a MOMENTUM = MASS * VELOCITY
From the above calculations, momentum on impact = 100 * sqrt (2*9.81*10) = 1400
This momentum is absorbed by the ground on which the mass impacted. If you knew the time (duration) of impact, you could calculate the force on the ground, given by Force = Change in Momentum / Duration of impact

Answer 4

Force is just 100 kg (kg force, not kg mass), it does not change by the speed. What changes is the kinetic energy of the block, that is

1/2 * m * v^2 = 0.5 * 100 * (2*9.8*10) = 9800 kg m^2/sec^2

This energy does a work deforming the ground. So you must not look for the force, but for the energy.

Answer 5

The formulation is real. The vertical tension appearing on the physique in loose fall is weight - assuming no drag - and weight = m*g. What you opt to comprehend is the fee of the physique earlier result and the time it takes to relax. At that factor you could make certain the stress of the effect by skill of F = m * (vf - vi) / t. or you could think of in terms of a bullet/block difficulty in which you comprehend ways a techniques the bullet is going into the block.. a = vi^2/(2*d) so F = m*vi^2/(2*d).