A steel safe with mass M falls onto a concrete floor. Just before hitting the floor, its speed is vi . It hits the floor without rebounding, and ends up being d shorter than before. Find the magnitude of the (average) force exerted by the floor on the safe while it is being deformed. You can assume this force is a constant during the short time it takes for the safe to deform. Ignore the effects of gravity.

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Answer:

Explanation:

initial velocity, u = vi

final velocity, v = 0

distance = d

mass of safe = M

Use third equation of motion

v² = u² + 2as

0 = vi² - 2 ad

a = vi²/2d

Force, F = mass x acceleration

F = M x vi²/2d

F = Mvi²/2d

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