Math Problem Statement

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?

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?

Solution

We are given the following data from the problem:

  • Mass of the block: m=2.4kgm = 2.4 \, \text{kg}
  • Spring constant: k=422N/mk = 422 \, \text{N/m}
  • Maximum compression of the spring: x=14.0cm=0.14mx = 14.0 \, \text{cm} = 0.14 \, \text{m}
  • Coefficient of kinetic friction: μk=0.44\mu_k = 0.44
  • Acceleration due to gravity: g=9.81m/s2g = 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 WsW_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:

Ws=12kx2W_s = \frac{1}{2} k x^2

Substituting the given values:

Ws=12×422N/m×(0.14m)2W_s = \frac{1}{2} \times 422 \, \text{N/m} \times (0.14 \, \text{m})^2

Ws=4.13468JW_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 WfW_f is given by:

Wf=fkd=μkmgdW_f = f_k \cdot d = \mu_k \cdot m \cdot g \cdot d

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

Wf=μkmgxW_f = \mu_k \cdot m \cdot g \cdot x

Substituting the given values:

Wf=0.44×2.4kg×9.81m/s2×0.14mW_f = 0.44 \times 2.4 \, \text{kg} \times 9.81 \, \text{m/s}^2 \times 0.14 \, \text{m}

Wf=1.4491JW_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:

12mv2=Ws+Wf\frac{1}{2} m v^2 = W_s + W_f

12×2.4kg×v2=4.13J+1.45J\frac{1}{2} \times 2.4 \, \text{kg} \times v^2 = 4.13 \, \text{J} + 1.45 \, \text{J}

1.2v2=5.58J1.2 \, v^2 = 5.58 \, \text{J}

Solving for vv:

v2=5.581.2v^2 = \frac{5.58}{1.2}

v=4.65v = \sqrt{4.65}

v2.16m/sv \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.

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Math Problem Analysis

Mathematical Concepts

Work-Energy Principle
Spring Potential Energy
Kinetic Friction
Kinematics

Formulas

W_s = 0.5 * k * x^2 (Work done by the spring)
W_f = μ_k * m * g * d (Work done by friction)
Kinetic energy: 0.5 * m * v^2 = W_s + W_f

Theorems

Work-Energy Theorem
Hooke's Law for Springs

Suitable Grade Level

Grades 11-12