Question #40306

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Question #40306

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Answer 1 · 2016-11-16T08:09:15

The mass of the man $= 76 k g$ $=76kg$
The downward gravitational force on him
$= 76 \times g N = 76 \times 9.8 N = 744.8 N$ $=76xxgN=76xx9.8N=744.8N$

The mass of the Parachute $= 5.5 k g$ $=5.5kg$
The downward gravitational force on it
$= 5.5 \times g N = 5.5 \times 9.8 N = 53.9 N$ $=5.5xxgN=5.5xx9.8N=53.9N$

So total downward gravitational pull
$F_{g} ⏐ ↓ = (744.8 + 53.9) N = 798.7 N$ $F_gdarr = (744.8+ 53.9)N = 798.7N$

The net downward acceleration on the system is $2.6 m s^{- 2}$ $2.6ms^-2$

So the net downward force

$F_{n} ⏐ ↓ = (76 + 5.5) \times 2.6 = 211.9 N$ $F_ndarr=(76+5.5)xx2.6=211.9N$

Now

$F_{n} ⏐ ↓ = F_{g} ⏐ ↓ - F_{air} ↑ ⏐$ $F_ndarr=F_gdarr-F_"air"uarr$

$\Rightarrow F_{air} ↑ ⏐ = F_{g} ⏐ ↓ - F_{n} ⏐ ↓$ $=>F_"air"uarr=F_gdarr-F_ndarr$

$\Rightarrow F_{air} ↑ = 798.7 - 211.9 = 586.8 N$ $=>F_"air"uarr=798.7-211.9=586.8N$

(a) The upward force exerted on the open parachute by the air $= F_{air} ↑ = 586.8 N$ $=F_"air"uarr=586.8N$

(b) The downward force exerted by the person on the parachute $= 744.8 N$ $=744.8N$

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Question #40306

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