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Force exerted to a falling object

  • Thread starter Perdition
  • Start date
  • #1
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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 im stumped on how do i find the forces exerted by earth to stop the rock from digging deeper.
 
Last edited:

Answers and Replies

  • #2
54
0
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
 
  • #3
CAF123
Gold Member
2,902
88
Think of the 20mm as a 'stopping distance'.
 
  • #4
5
0
i assume you substitute s with 0.006 m right. I tried that in that case

[itex]v^{2}_{f}[/itex]=[itex]v^{2}_{i}[/itex] + 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
 
Last edited:
  • #5
54
0
use mass = 120/9.8

it gives an approximate answer
 
  • #6
5
0
getting closer :P thanks I guess it was just an error in rounding the number
 

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