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Sagot :
When a mass of 3 kg is hanged to it, the string will stretch 7.5 cm.
Given,
The length of the spring attached to the ceiling = 10 cm
The quantity of the mass hanged in the spring = 2 kg
The amount of stretching happened in the spring = 15 cm
We have to find the stretching of string when a mass of 3 kg is hanged to it;
Here,
Given that the force applied to the spring by the 2.0 kg is equal to the mass's weight:
F = mg = 2 kg × 9.8 m/s² = 19.6 N
Additionally, the stretching of:
Δx = 15 cm - 10 cm = 5 cm
So, applying Hooke's law:
F = kΔx
To determine the spring constant, k;
k = F/Δx = 19.6/5 = 3.92 N/ cm
The spring is now coupled to a new mass of m = 3.0 kg; the force this mass exerts on the spring equals
F = mg = 3 kg × 9.8 m/s² = 29.4 N
Thus, we may apply Hooke's law once more to determine the new stretching:
Δx = F/k = 29.4/3.92 = 7.5 cm
Therefore,
The string will be stretched 7.5 cm when a mass of 3 kg is hanged to it.
Learn more about stretching of string here;
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