You push a 1.9 kg block against a horizontal spring, compressing the spring by 13 cm. Then you release the block, and the spring sends it sliding across a tabletop. It stops 60 cm from where you released it. The spring constant is 200 N/m. What is the block-table coefficient of kinetic friction?
To find the block-table coefficient of kinetic friction, we need to determine the acceleration of the block as it slides across the tabletop. We can use the concept of conservation of mechanical energy to solve this problem.
1. First, let's find the potential energy stored in the compressed spring. The potential energy of the spring is given by the formula: \(PE = \frac{1}{2} k x^2\), where \(k\) is the spring constant and \(x\) is the displacement from the equilibrium position in meters.
Plugging in the given values, we have:
\(PE = \frac{1}{2} \times 200 \, \text{N/m} \times (0.13 \, \text{m})^2 = 1.055 \, \text{J}\).
2. According to the law of conservation of mechanical energy, this potential energy is converted into kinetic energy as the block slides across the tabletop. The kinetic energy of a moving object is given by the formula: \(KE = \frac{1}{2} m v^2\), where \(m\) is the mass of the object and \(v\) is its velocity in m/s.
Since the block stops eventually, we know that all of the initial potential energy is converted into kinetic energy up to that point. So we can express the equation as:
\(1.055 \, \text{J} = \frac{1}{2} \times 1.9 \, \text{kg} \times v^2\).
3. Solving for \(v\), we have:
\(v^2 = \frac{2 \times 1.055 \, \text{J}}{1.9 \, \text{kg}}\).
\(v^2 = 1.11 \, \text{m}^2/\text{s}^2\).
4. Now, we can determine the acceleration of the block using the equation: \(v^2 = u^2 + 2a s\), where \(u\) is the initial velocity (which is zero since the block was released from rest) and \(s\) is the distance traveled.
Rearranging the equation, we have:
\(a = \frac{v^2}{2s}\).
\(a = \frac{1.11 \, \text{m}^2/\text{s}^2}{2 \times 0.6 \, \text{m}}\).
\(a = 0.925 \, \text{m/s}^2\).
5. Finally, to find the block-table coefficient of kinetic friction, we can use the equation: \(f_k = \mu_k \times m \times g\), where \(f_k\) is the force of kinetic friction and \(g\) is the acceleration due to gravity.
Since the block is in equilibrium, the force of kinetic friction is equal to the mass of the block multiplied by the acceleration, but with an opposite sign:
\(f_k = -m \times a\).
\(\mu_k \times m \times g = -m \times a\).
\(\mu_k = \frac{-m \times a}{m \times g}\).
\(\mu_k = -\frac{a}{g}\).
\(\mu_k = -\frac{0.925 \, \text{m/s}^2}{9.8 \, \text{m/s}^2}\).
After evaluating the expression, we find:
\(\mu_k \approx -0.094\).
Since the coefficient of kinetic friction cannot be negative, we take the absolute value, and the block-table coefficient of kinetic friction is approximately 0.094.