Force exerted to a falling object

AI Thread Summary
A rock weighing 120 N falls from a height of 2.00 m and penetrates the Earth by 60 mm upon impact. The velocity of the rock just before collision is calculated to be 6.3 m/s using conservation of energy. To find the Earth's average force of resistance, the acceleration needed to stop the rock is determined using the equation v² = v₀² + 2as, leading to an acceleration of approximately 330.75 m/s². Multiplying this acceleration by the mass (12.24 kg) results in a force of about 3969 N, which is close to the expected answer of 4.1 kN after adjusting for rounding. The discussion highlights the importance of accurate calculations and unit conversions in physics problems.
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Homework Statement



A rock with a weight of 120 N free falls from height of 2.00m collides with with Earth and digs 60 mm into it. What is the Earth's average force of resistance

Homework Equations



F=m*g => m=F/g

m*g*h = 1/2*m*v^2

v=√(2*g*h)

The Attempt at a Solution



I found the velocity of a rock just before collision using the law of conservation of energy. I now know the velocity which I got 6.3 m/s but now I am stumped on how do i find the forces exerted by Earth to stop the rock from digging deeper.
 
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Hmm i would do it another way

Before hitting ground its velocity is 6.3m/s right??

So that velocity becomes 0 due to some acceleration (due to resistance force)

so using v2-v02=2(a)(s)

u find acceleration (retardation in this case)

now that multiplied to the mass will give u the force...

Hope it helps
 
Think of the 20mm as a 'stopping distance'.
 
i assume you substitute s with 0.006 m right. I tried that in that case

v^{2}_{f}=v^{2}_{i} + 2 * a * x

0=6.3^{2}+2*a*0.06
0=39.69+0.12a
-0.012a=39.69 / (-0.12a)
a=330.75 m/s^2

F=m*a

F= 12 * 330.75 = 3969 N

however the answer should be 4.1 kN
 
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use mass = 120/9.8

it gives an approximate answer
 
getting closer :P thanks I guess it was just an error in rounding the number
 
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