Physics, asked by geetyanium, 1 year ago


A stone of mass 5kg falls from the top of a cliff 40m and buries itself 2m in ground.Find the average force offered by sand and the time it takes to penetrate.g is 9.8 ms-2

Answers

Answered by kvnmurty
3
Velocity of the stone just before striking the ground = v
v² = u² + 2 g s
v² = 0 + 2* 9.8 * 40 = 49 * 4
=> v = 14 m/s

Let the deceleration of the stone inside ground be a.
v² = u² + 2 a s
0 = 14² + 2 a * 2
=> a = - 49 m/s²

Time taken to penetrate inside ground and stop = t
 t =  (v - u)/a = (0 - 14)/(-49) = 2/7 sec.

Avg Force offered by the Sand = m a = 5 kg * (-49) m/s²
   = - 245 Newtons
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