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A moving 2.4 kg block collides with a horizontal spring whose spring constant is 422 N/m (see figure). The block compresses the spring a maximum distance of 14.0 cm from its rest position. The coefficient of kinetic friction between the block and the horizontal surface is 0.44. What is the work done by the spring in bringing the block to rest?

How much mechanical energy is being dissipated by the force of friction while the block is being brought to rest by the spring?

What is the speed of the block when it hits the spring?

We are given the following data from the problem:

- Mass of the block: \( m = 2.4 \, \text{kg} \)
- Spring constant: \( k = 422 \, \text{N/m} \)
- Maximum compression of the spring: \( x = 14.0 \, \text{cm} = 0.14 \, \text{m} \)
- Coefficient of kinetic friction: \( \mu_k = 0.44 \)
- Acceleration due to gravity: \( g = 9.81 \, \text{m/s}^2 \)

### Part 1: Work done by the spring in bringing the block to rest

The work done by the spring \( W_s \) is equal to the potential energy stored in the spring when it is maximally compressed. The potential energy stored in a spring is given by:

\[
W_s = \frac{1}{2} k x^2
\]

Substituting the given values:

\[
W_s = \frac{1}{2} \times 422 \, \text{N/m} \times (0.14 \, \text{m})^2
\]

\[
W_s = 4.13468 \, \text{J}
\]

Thus, the work done by the spring is approximately **4.13 J**.

### Part 2: Mechanical energy dissipated by friction

The work done by friction \( W_f \) is given by:

\[
W_f = f_k \cdot d = \mu_k \cdot m \cdot g \cdot d
\]

Here, \( f_k = \mu_k \cdot m \cdot g \) is the force of kinetic friction, and \( d \) is the distance the block moves while compressing the spring (which is the same as the spring compression \( x \)).

\[
W_f = \mu_k \cdot m \cdot g \cdot x
\]

Substituting the given values:

\[
W_f = 0.44 \times 2.4 \, \text{kg} \times 9.81 \, \text{m/s}^2 \times 0.14 \, \text{m}
\]

\[
W_f = 1.4491 \, \text{J}
\]

Thus, the mechanical energy dissipated by friction is approximately **1.45 J**.

### Part 3: Speed of the block when it hits the spring

To find the speed of the block when it hits the spring, we use the work-energy principle. The total mechanical energy initially stored in the block's kinetic energy must equal the work done by both the spring and friction. Thus:

\[
\frac{1}{2} m v^2 = W_s + W_f
\]

\[
\frac{1}{2} \times 2.4 \, \text{kg} \times v^2 = 4.13 \, \text{J} + 1.45 \, \text{J}
\]

\[
1.2 \, v^2 = 5.58 \, \text{J}
\]

Solving for \( v \):

\[
v^2 = \frac{5.58}{1.2}
\]

\[
v = \sqrt{4.65}
\]

\[
v \approx 2.16 \, \text{m/s}
\]

Thus, the speed of the block when it hits the spring is approximately **2.16 m/s**.

---

Would you like further details or have any questions?

### Follow-up Questions:
1. How does the spring constant affect the maximum compression?
2. What would happen if the coefficient of friction were higher?
3. How could we calculate the time it takes for the block to stop?
4. How would the system behave if there were no friction?
5. Can we calculate the deceleration of the block due to friction?

### Tip:
Always double-check units when working with energy and force equations, as consistency in units is key to obtaining the correct result.

A moving 2.4 kg block collides with a horizontal spring whose spring constant is 422 N/m (see figure). The block compresses the spring a maximum distance of 14.0 cm from its rest position. The coefficient of kinetic friction between the block and the horizontal surface is 0.44. What is the work done by the spring in bringing the block to rest? How much mechanical energy is being dissipated by the force of friction while the block is being brought to rest by the spring? What is the speed of the block when it hits the spring?

Find out how to calculate the work done by a spring with a constant of 422 N/m as a 2.4 kg block compresses it by 14 cm. This problem explores key concepts like spring potential energy, friction, and the work-energy principle. Ideal for high school physics students.

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