Answer :
The spring is stretched by 0.103 m from its unstrained length.
To find the amount by which the spring is stretched, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position. The formula for Hooke's Law is given by F = -kx, where F is the force exerted by the spring, k is the spring constant, and x is the displacement from the equilibrium position. In this case, we can rearrange the formula to solve for x: x = -F/k.
Given that the mass of the object is 1.2 kg and the spring constant is 114 N/m, we can calculate the force exerted by the spring using the formula F = mg, where m is the mass of the object and g is the acceleration due to gravity. Plugging in the values, we get F = 1.2 kg * 9.8 m/s^2 = 11.76 N.
Now we can use the formula x = -F/k to calculate the displacement of the spring: x = -11.76 N / 114 N/m = -0.103 m. Since displacement is a vector quantity, we take the absolute value to get the magnitude of the displacement, which in this case is 0.103 m. Therefore, the spring is stretched by 0.103 m from its unstrained length.
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