A spring ( k =805N/m) is hangingfrom the ceiling of an elevator, and a 4.1-kg ob

Question

A spring (k =805N/m) is hangingfrom the ceiling of an elevator, and a 4.1-kg object is attached tothe lower end. By how much does the spring stretch (relative to itsunstrained length) when the elevator is accelerating upward ata = 0.24m/s2? Thank You!!!
A spring (k =805N/m) is hangingfrom the ceiling of an elevator, and a 4.1-kg object is attached tothe lower end. By how much does the spring stretch (relative to itsunstrained length) when the elevator is accelerating upward ata = 0.24m/s2? Thank You!!!
Thank You!!!

Explanation / Answer

Apperent weight W = m ( a + g )                                = 4.1 ( 0.24 +9.8 ) = 41.164 N we know at equilibrium , kx = W from this spring streatchness x = W / k                                               = 41.164 N / 805 N / m                                                = 0.0511 m

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