A block on a frictionless, horizontal surface is attached to a spring that requires 125 N of force for the block to be stretched 15 cm away from equilibrium. From this position the block is released, and it begins to oscillate with a period of 0.25 s. What is the mass of the block?
A) 0.66 kg
B) 4.1 kg
C) 1.3 kg
D) 5.3 kg

Question

Quizplus · Accepted Answer

Using the formula for the period of a spring-mass system, we have:
T = 2π√(m/k)
where T = 0.25 s, k = (125 N) / (0.15 m) = 833.33 N/m (since F = kx), and we are solving for m.

Plugging in the values, we get:
0.25 s = 2π√(m/833.33 N/m)
Simplifying:
0.25 s = 2π√(m/(833.33 kg/s^2))
0.25 s = 2π√(m/833.33) s^-1
0.25 s = 2π(0.0469)√m
√m = 0.25 s / (2π(0.0469))
√m ≈ 0.663
m ≈ (0.663)^2 ≈ 0.440 kg

Therefore, the closest answer choice is C) 1.3 kg.

A Block on a Frictionless, Horizontal Surface Is Attached to a Spring