A 4.55 kg flowerpot drops from a tall building. (a) What is the magnitude of the

Question

A 4.55 kg flowerpot drops from a tall building.

(a) What is the magnitude of the force acting on the pot while it is in the air, 1.50 s after it begins to fall?

(b) After the pot has fallen 21.5 m, what is its speed?

(c) After the pot has fallen 21.5 m, it enters a viscous liquid, which brings it to rest over a distance of

1.50 m. Assuming constant deceleration over this distance, what is the magnitude of this deceleration?

(d) What is the magnitude of the force exerted on the bot by the liquid?

Explanation / Answer

(a) force is always mg while its in air.. that is mg

mg = 4,55*9.8 = 44.59 N

(b) v^2 = u^2 +2as

u = 0

a = 9.8

s = 21.5

so u get

v = 20.53 m/s

(c) change in velocity = 20.53 m/s

distance = 1.5 m

(20.53)^2 = 2a*1.5

a = 140.47 m/s2

(d) F = ma = 4.55*140.47 = 639.12 N

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