QUESTION IMAGE
Question
overview
- in a few sentences, write down the overall objective of this lab. (5 pts.)
- explain why the coefficient of kinetic friction, μk, is independent of the tension in the string, t. (3 pts.)
- which would result in a stronger force of static friction, μs = 0.36 or μs = 0.42? (2 pts.)
data (43 pts.)
ma = 5g
| experiment 1 | plastic | felt | cork | |
|---|---|---|---|---|
| a1 | 2.24 | 1.46 | 0.605 | |
| a2 | 2.16 | 1.84 | 0.889 | +5g |
| a3 | 2.15 | 1.08 | 0.756 | |
| a4 | 2.24 | 1.47 | 0.730 | |
| a5 | 2.14 | 1.71 | 0.526 | |
| a6 | 2.24 | 1.47 | 0.914 | |
| aave | 2.20 | 1.58 | 0.71 | |
| μk |
Question 1
The overall objective of the lab cannot be determined precisely without more context about the lab. However, it likely involves studying friction - related concepts as evidenced by the questions about coefficients of friction.
Question 2
The coefficient of kinetic friction $\mu_k$ depends on the nature of the two surfaces in contact. It is a property of the materials and their surface conditions. Tension in a string affects the net - force acting on an object and its motion, but not the microscopic interactions between the surfaces that determine $\mu_k$. Mathematically, $F_f=\mu_k N$, where $F_f$ is the frictional force, $N$ is the normal force, and $\mu_k$ is independent of any external forces like tension in the string as long as the surfaces remain the same.
Question 3
The force of static friction is given by $F_{s,max}=\mu_s N$, where $\mu_s$ is the coefficient of static friction and $N$ is the normal force. Assuming the normal force $N$ is the same in both cases, a larger coefficient of static friction results in a larger maximum static - friction force. Since $0.42>0.36$, $\mu_s = 0.42$ will result in a stronger force of static friction.
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- Without more context, the overall objective of the lab is likely related to studying friction concepts.
- The coefficient of kinetic friction depends on surface properties, not tension.
- $\mu_s = 0.42$ results in a stronger force of static friction.