Math Problem Statement

Problem 2 An ideal spring is hung vertically from the ceiling. When a 2.0-kg mass hangs at rest from it, the spring is extended 6.0 cm from its relaxed length. An upward external force is then applied to the block to move it upward a distance of 16 cm. While the block is being raised by the force, the work done by the spring is

Solution

To solve this problem, we need to calculate the work done by the spring as it is stretched or compressed.

Given:

  • Mass, m=2.0kgm = 2.0 \, \text{kg}
  • Extension due to the mass, Δx1=6.0cm=0.06m\Delta x_1 = 6.0 \, \text{cm} = 0.06 \, \text{m}
  • The distance the block is raised, Δx2=16.0cm=0.16m\Delta x_2 = 16.0 \, \text{cm} = 0.16 \, \text{m}

Step 1: Determine the spring constant, kk

From Hooke's law: F=kΔx1F = k \Delta x_1 The force applied by the weight of the mass when at rest is the gravitational force: F=mg=2.0kg×9.8m/s2=19.6NF = m g = 2.0 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 19.6 \, \text{N} Thus, k=FΔx1=19.6N0.06m=326.67N/mk = \frac{F}{\Delta x_1} = \frac{19.6 \, \text{N}}{0.06 \, \text{m}} = 326.67 \, \text{N/m}

Step 2: Calculate the initial and final displacements

  • Initial displacement from equilibrium: x1=0.06mx_1 = 0.06 \, \text{m}
  • Final displacement after being raised: x2=0.06m0.16m=0.10mx_2 = 0.06 \, \text{m} - 0.16 \, \text{m} = -0.10 \, \text{m} (since the block moves upward, reducing the displacement).

Step 3: Work done by the spring

The work done by a spring is given by: Wspring=12k(x12x22)W_{\text{spring}} = \frac{1}{2} k \left( x_1^2 - x_2^2 \right) Substitute the known values: Wspring=12×326.67N/m×((0.06m)2(0.10m)2)W_{\text{spring}} = \frac{1}{2} \times 326.67 \, \text{N/m} \times \left( (0.06 \, \text{m})^2 - (-0.10 \, \text{m})^2 \right) Wspring=163.33×(0.00360.01)W_{\text{spring}} = 163.33 \times \left( 0.0036 - 0.01 \right) Wspring=163.33×(0.0064)W_{\text{spring}} = 163.33 \times (-0.0064) Wspring=1.045JW_{\text{spring}} = -1.045 \, \text{J}

Final Answer:

The work done by the spring is 1.045J-1.045 \, \text{J}. The negative sign indicates that the spring is doing negative work, which means it is resisting the upward motion of the block.

Would you like further explanation on any part of this solution?

Related Questions:

  1. How does Hooke's law apply to spring systems?
  2. What is the significance of negative work in a mechanical system?
  3. How does changing the mass affect the spring's extension in this scenario?
  4. How can the potential energy of a spring be related to the work done?
  5. What is the impact of gravitational force when considering the spring's work?

Tip: Always pay attention to the direction of displacement when calculating work, as it affects whether the work is positive or negative.

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Math Problem Analysis

Mathematical Concepts

Hooke's Law
Work-Energy Principle
Kinematics

Formulas

F = kΔx (Hooke's Law)
W_spring = (1/2)k(x1^2 - x2^2) (Work done by a spring)

Theorems

Hooke's Law: Force is proportional to displacement in a spring system.
Work-Energy Principle: The work done on an object relates to its energy change.

Suitable Grade Level

Grades 11-12 (High School Physics)