A stone of masss 5 kg falls from the top of a cliff 50m height and burries 1m deep in sand.Find the average resistance offered by the sand and the time it takes to penetrate.
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If velocity of stone when reaches the ground is v then
v² - u² = 2aS
v² = 2gH
= 2 × 9.8 m/s² × 50 m
= 980 m²/s²
a = -v² / (2S)
= -[980 m²/s²] / (2 × 1 m)
= -490 m/s²
Here negative sign indicates that stone is undergoing retardation inside the soil.
F = ma
= 5 kg × (-490 m/s²)
= -2450 N
Here negative sign indicates that Force is resistance.
Average resistance offered is 2450 N
v = u + at
0 = u + at
t = u / (-a)
t = sqrt(980 m/s) / (-(-490 m/s))
t = 0.0638 seconds
Time taken by it to penetrate is 0.0638 seconds.
v² - u² = 2aS
v² = 2gH
= 2 × 9.8 m/s² × 50 m
= 980 m²/s²
a = -v² / (2S)
= -[980 m²/s²] / (2 × 1 m)
= -490 m/s²
Here negative sign indicates that stone is undergoing retardation inside the soil.
F = ma
= 5 kg × (-490 m/s²)
= -2450 N
Here negative sign indicates that Force is resistance.
Average resistance offered is 2450 N
v = u + at
0 = u + at
t = u / (-a)
t = sqrt(980 m/s) / (-(-490 m/s))
t = 0.0638 seconds
Time taken by it to penetrate is 0.0638 seconds.
Kattarhindu:
ok right but also find the time
Oh. Sorry. I forgot. I'll edit my answer soon.
yes
Answer eidited.
great
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