When a 3.30-kg object is hung vertically on a certain light spring that obeys Hooke's law, the spring stretches 2.80 cm. If the 3.30-kg object is removed, how far will the spring stretch if a 1.50-kg block is hung on it?

0.0127 m or 1.27 cm

Explanation:

From Hook's law,

F = ke Equation 1

Where F = Force on the spring, k = spring constant of the spring, e = extension of the spring.

Note: The Force of the spring is the weight of the mass hung on the spring,

I.e,

F = mg Equation 2

Where m = mass of the object hung on the spring, g = acceleration due to gravity

Substitute equation 2 into equation 1

mg = ke Equation 3

make k the subject of the equation

k = mg/e Equation 4

Given: m = 3.3 kg,  e = 2.8 cm = 0.028 m, g = 9.8 m/s²

Substitute into equation 4

k = 3.3(9.8)/0.028

k = 1155 N/m

When the 3.3 kg block is removed,

make e the subject of the equation

e = mg/k Equation 5

Given: m = 1.5 kg, g = 9.8 m/s², k = 1155 N/m

Substitute into equation 5

e = 1.5(9.8)/1155

e = 0.0127 m

e = 1.27 cm

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