What is the value of a spring constant when a 100g mass is attached to a spring and it is accelerated 0.5 ms-2 through a displacement of 5cm?
Correct answer: C. 1 Nm-1
- A. 0.1 Nm-1
- B. 0.5 Nm-1
- C. 1 Nm-1
- D. 5 Nm-1
Explanation
To find the value of the spring constant (k), 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 = -k * x Where: F = Force exerted by the spring k = Spring constant x = Displacement from the equilibrium position In this case, the spring is accelerating a 100g mass (0.1 kg) with an acceleration of 0.5 m/s² over a displacement of 5 cm (0.05 m). First, let's find the force (F) required to accelerate the mass: F = m * a Where: m = Mass (0.1 kg) a = Acceleration (0.5 m/s²) F = 0.1 kg * 0.5 m/s² F = 0.05 kg m/s² = 0.05 N Now, using Hooke's Law, we can find the spring constant (k): F = -k * x 0.05 N = -k * 0.05 m To isolate k, divide both sides by -0.05 m: k = 0.05 N / 0.05 m k = 1 N/m So, the value of the spring constant (k) is 1 N/m.
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