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If you hang a body with a mass of 0.01kg on a vertical spring and it stretches by 4.8mm, what is the elastic constant?

User Itay Oded
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Final answer:

To find the elastic constant of the spring, 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. In this case, the elastic constant is approximately 20.42 N/m.

Step-by-step explanation:

To find the elastic constant of the spring, 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 F = kx, where F is the force exerted by the spring, k is the elastic constant (also known as the spring constant), and x is the displacement of the spring.

In this case, we're given a mass of 0.01 kg and a displacement of 4.8 mm (or 0.0048 m). Since the force is equal to the weight of the mass hanging on the spring, we can use the formula F = mg, where m is the mass and g is the acceleration due to gravity (approximately 9.8 m/s²).

Substituting the given values into the equation, we have F = (0.01 kg)(9.8 m/s²) = 0.098 N. Setting this equal to kx, we have 0.098 N = k(0.0048 m). Solving for k, we find that the elastic constant of the spring is approximately 20.42 N/m.

User James Murty
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