Answer:
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
The extension of the spring in the elevator is 60 mm.
For the extension of the spring to be zero, the elevator must be moving downwards under free fall.
The given parameters;
The spring constant is calculated as follows;
F = kx
mg = kx
[tex]k = \frac{mg}{x} \\\\k = \frac{5 \times 9.8}{0.05} \\\\k = 980 \ N/m[/tex]
The tension on the spring in an elevator accelerating upwards is calculated as follows;
T = mg + ma
T = m(g + a)
T = 5(9.8 + 2)
T = 59 N
The extension of the spring is calculated as follows;
[tex]T = kx\\\\x = \frac{T}{k} \\\\x = \frac{59}{980} \\\\x = 0.06 \ m\\\\x = 60 \ mm[/tex]
For the extension of the spring to be zero, the elevator must be under free fall, such that the tension on the spring is zero.
For free fall, a = g
T = m(g - a) = 0
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