A 77 kg man drops to a concrete patio from a window 0.55 m above the patio. He n

Question

A 77 kg man drops to a concrete patio from a window 0.55 m above the patio. He neglects to bend his knees on landing, taking 2.4 cm to stop.

(a) What is his average acceleration from when his feet first touch the patio to when he stops?

(b) What is the magnitude of the average stopping force exerted on him by the patio?

A 77 kg man drops to a concrete patio from a window 0.55 m above the patio. He neglects to bend his knees on landing, taking 2.4 cm to stop (a) What is his average acceleration from when his feet first touch the patio to when he stops? 224.133 m/s2 (b) What is the magnitude of the average stopping force exerted on him by the patio? XkN Fnter a nu

Explanation / Answer

landing velocity v² = u² + 2gh

(u=0)

v² = 2gh = 2* 9.8*0.55

v² = 10.78 m/s²

stopping deceleration (-a)

v² = u² - 2as

(v = 0, u² = 10.791m/s², s=0.02m)

a = u² / 2s

a = 10.78 / (2*0.024)

a = -224.58 m/s²

-a = 224.58 m/s²

Force = m*a

F = 77 kg*224.58

F = 17292.66 N

F = 17.3 kN

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