A 620.0 g block of wood is firmly attached to a light horizontal spring with an elastic constant of 180.0 N/m. When the system is compressed 5.0 cm and released, it elongates 2.3 cm beyond the equilibrium position before stopping and returning. What is the coefficient of kinetic friction of the floor? Write your answer to two decimal places.
The correct answer and explanation is:
To find the coefficient of kinetic friction (μₖ), we can use the work-energy principle. The spring system starts compressed, moves through equilibrium, and stretches beyond it before stopping. The energy lost is due to friction, so:
Given:
- Mass of the block, m = 620.0 g = 0.620 kg
- Spring constant, k = 180.0 N/m
- Initial compression from equilibrium, x₁ = 5.0 cm = 0.050 m
- Elongation beyond equilibrium, x₂ = 2.3 cm = 0.023 m
- Total distance moved, d = x₁ + x₂ = 0.050 + 0.023 = 0.073 m
Step 1: Initial Energy (Spring Potential Energy)
When the spring is compressed 5.0 cm, the initial potential energy stored is: PE=12kx12=12(180.0)(0.050)2=0.225 JPE = \frac{1}{2} k x_1^2 = \frac{1}{2} (180.0) (0.050)^2 = 0.225 \, \text{J}
Step 2: Final Energy
At the maximum stretch (2.3 cm beyond equilibrium), the final potential energy is: PE=12kx22=12(180.0)(0.023)2=0.04761 JPE = \frac{1}{2} k x_2^2 = \frac{1}{2} (180.0) (0.023)^2 = 0.04761 \, \text{J}
Step 3: Work Done by Friction
The work done by friction is equal to the initial PE − final PE: Wfriction=0.225−0.04761=0.17739 JW_{\text{friction}} = 0.225 – 0.04761 = 0.17739 \, \text{J}
Step 4: Friction Force
fk=μk⋅N=μk⋅(mg)=μk⋅(0.620⋅9.8)=μk⋅6.076f_k = \mu_k \cdot N = \mu_k \cdot (m g) = \mu_k \cdot (0.620 \cdot 9.8) = \mu_k \cdot 6.076
Step 5: Work by Friction
Wfriction=fk⋅d=μk⋅6.076⋅0.073=0.443548μkW_{\text{friction}} = f_k \cdot d = \mu_k \cdot 6.076 \cdot 0.073 = 0.443548 \mu_k
Set this equal to 0.17739 J: 0.443548μk=0.177390.443548 \mu_k = 0.17739
Solve for μₖ: μk=0.177390.443548≈0.40\mu_k = \frac{0.17739}{0.443548} \approx 0.40
✅ Final Answer:
Coefficient of kinetic friction μₖ = 0.40
🧠 Explanation (300 words):
This problem uses the work-energy principle to find the coefficient of kinetic friction between a wooden block and a floor. When a spring is compressed and then released, it converts stored elastic potential energy into kinetic energy, which moves the block across the surface. However, because of friction, not all of that energy is conserved. Some is lost as thermal energy due to sliding.
Initially, when the spring is compressed, it holds a certain amount of potential energy based on the formula 12kx2\frac{1}{2}kx^2. As the spring expands, that energy is transformed into motion. However, since the block doesn’t return all the way symmetrically, we can conclude that friction is opposing the motion.
By comparing the initial energy (compressed) and the final energy (after elongation), we can calculate how much energy was lost due to friction. Then, using the formula for work (W=f⋅dW = f \cdot d), we determine the frictional force. Since friction is also equal to the coefficient of kinetic friction multiplied by the normal force (which equals weight here), we solve for μₖ.
This process elegantly demonstrates energy conservation in the presence of a non-conservative force like friction.