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Attempts 127

A 5-kg block is suspended from a spring, causing the spring to stretch 10 cm from equilibrium. What is the spring constant for this spring?

  1. A
    4.9 N/cm
  2. B
    9.8 N/cm
  3. C
    49 N/cm
  4. D
    50 N/cm

Topic Flashcards

Click to Flip
Question

State Hooke's Law for an ideal spring in equation form, and define each variable.

Answer

F = -k * x, where F is the restoring force, k is the spring constant (stiffness), and x is the displacement from equilibrium. (The negative sign indicates the force opposes the displacement.)

Question

In a static equilibrium setup with a hanging mass, what force balances the spring's restoring force?

Answer

The weight of the hanging mass (mg).

Question

For the 5-kg block, calculate the weight force that stretches the spring. (Use g = 9.8 m/s²)

Answer

F = m * g = 5 kg * 9.8 m/s² = 49 Newtons (N).

Question

How do you calculate the spring constant (k) from a known force (F) and the resulting stretch (x)?

Answer

k = F / x

Question

If the same 5-kg block stretched a different spring by 20 cm, would its spring constant be higher or lower than in the original problem?

Answer

Lower. A larger stretch for the same force means a less stiff (smaller k) spring. (k = 49 N / 0.2 m = 245 N/m or 2.45 N/cm).

Mini Quiz

1 / 3
In the equilibrium position, is the net force on the suspended block zero?

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