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Brums [2.3K]
3 years ago
8

A spring 1.7 m long with force constant 550 N/m is hung from the ceiling of an elevator, and a block of mass 15 kg is attached t

o the bottom of the spring. (a) By how much is the spring stretched when the block is slowly lowered to its equilibrium point
Physics
1 answer:
Nastasia [14]3 years ago
3 0

Answer:

27 cm

Explanation:

ASSUMING the elevator is in a constant velocity state (could be zero)

mg = kx

x = mg/k = 15(9.8)/550 = 0.267272... m ≈ 27 cm

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Answer:

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In all cases which a body is on a ramp force N will be smaller than mg, and depending on the level of the ramp (bigger the ramp smaller force N)

Explanation:

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Therefore we can say that force N is always minor than mg unless in extreme case where the block is over the surface in which case N takes its maximum value N = mg

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