When a 5.0 kg box is hung from a spring, the spring stretches to 50 mm beyond its relaxed length. (a) In an elevator accelerating upward at 2.0 m/s2 , how far does the spring stretch with the same box attached? (b) How fast and in which direction should the elevator accelerate for the spring stretch to be zero (that is, the spring returns to its relaxed length)?

a) the spring will stretch 60.19 mm with the same box attached as it accelerates upwards

b) spring will be relaxed when the elevator accelerates downwards at 9.81 m/s²

Explanation:

Given that;

Gravitational acceleration g = 9.81 m/s²

Mass m = 5 kg

Extension of the spring X = 50 mm = 0.05 m

Spring constant k = ?

 we know that;

mg = kX  

5 × 9.81 = k(0.05)

k = 981 N/m

a)

Given that; Acceleration of the elevator a = 2 m/s² upwards

Extension of the spring in this situation = X1

Force exerted by the spring = F

we know that;

ma = F - mg

ma = kX1 - mg

we substitute

5 × 2 = 981 × X1 - (5 ×9.81 )

X1 = 0.06019 m

X1 = 60.19 mm

Therefore the spring  will stretch 60.19 mm with the same box attached as it accelerates upwards

B)

Acceleration of the elevator = a

The spring is relaxed i.e, it is not exerting any force on the box.

Only the weight force of the box is exerted on the box.

ma = mg

a = g

a = 9.81 m/s² downwards.

Therefore spring will be relaxed when the elevator accelerates downwards at 9.81 m/s²

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