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
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velocity v of the stone when it reaches the ground
v² = 2 h g = 2 * 40 * 9.8
=> v = 28 m/s
momentum = m v = 5 * 28 = 140 kg-m/s
The deceleration of the stone inside sand = - v²/2s
= - a = 28²/(2*2) = -14 m/s²
Force offered by sand = m a = - 70 N
Time taken to stop = t = -v/a = 28/14 = 2 sec
v² = 2 h g = 2 * 40 * 9.8
=> v = 28 m/s
momentum = m v = 5 * 28 = 140 kg-m/s
The deceleration of the stone inside sand = - v²/2s
= - a = 28²/(2*2) = -14 m/s²
Force offered by sand = m a = - 70 N
Time taken to stop = t = -v/a = 28/14 = 2 sec
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