You push a 2.0 kg block against a horizontal spring , compressing the spring by 12 cm. Then You release the block, and the spring sends it sliding across a tabletop. It stops 75 cm from where you released it. The spring constant is 170 N/m. What is the block-table coefficient of kinetic friction ?

Answer

To find the coefficient of kinetic friction (μ_k) between the block and the tabletop, we can follow these steps:

### Step 1: Calculate the potential energy stored in the spring
The potential energy (PE) stored in a compressed spring can be calculated using the formula:
\[ PE = \frac{1}{2} k x^2 \]
where:
- \( k = 170 \, \text{N/m} \) (spring constant)
- \( x = 0.12 \, \text{m} \) (compression of the spring)

Substituting the values:
\[ PE = \frac{1}{2} \times 170 \, \text{N/m} \times (0.12 \, \text{m})^2 \]
\[ PE = \frac{1}{2} \times 170 \times 0.0144 \]
\[ PE = \frac{1}{2} \times 2.448 \, \text{J} \]
\[ PE = 1.224 \, \text{J} \]

### Step 2: Calculate the work done by friction
The work done by friction (W_f) can be expressed as:
\[ W_f = F_f \times d \]
where:
- \( F_f = \mu_k \times m \times g \) (frictional force)
- \( d = 0.75 \, \text{m} \) (distance the block slides)

Substituting the expression for frictional force:
\[ W_f = \mu_k \times m \times g \times d \]
where:
- \( m = 2 \, \text{kg} \) (mass of the block)
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity)

So,
\[ W_f = \mu_k \times 2 \, \text{kg} \times 10 \, \text{m/s}^2 \times 0.75 \, \text{m} \]
\[ W_f = \mu_k \times 15 \, \text{N} \]

### Step 3: Set the work done by friction equal to the potential energy
Since all the potential energy from the spring is converted into work done against friction, we can set them equal:
\[ \mu_k \times 15 = 1.224 \]

### Step 4: Solve for μ_k
Now, we can solve for the coefficient of kinetic friction (μ_k):
\[ \mu_k = \frac{1.224}{15} \]
\[ \mu_k = 0.0816 \]

### Final Answer
The coefficient of kinetic friction (μ_k) between the block and the tabletop is approximately:
\[ \mu_k \approx 0.08 \]

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