Problem 65 - University Physics with Modern Physics

Question

You are an industrial engineer with a shipping company. As part of the package-handling system, a small box with mass 1.60 kg is placed against a light spring that is compressed 0.280 m. The spring has force constant \(k = 45.0\) N/m. The spring and box are released from rest, and the box travels along a horizontal surface for which the coefficient of kinetic friction with the box is \(\mu_k = 0.300\). When the box has traveled 0.280 m and the spring has reached its equilibrium length, the box loses contact with the spring. (a) What is the speed of the box at the instant when it leaves the spring? (b) What is the maximum speed of the box during its motion?

Step-by-Step Solution

Answer

(a) The speed is approximately 0.747 m/s. (b) The maximum speed is approximately 1.485 m/s.

Step 1: Calculate Spring Potential Energy (SPE)

Firstly, we need to calculate the potential energy stored in the spring when compressed. The formula for the spring potential energy is given as: \[ SPE = \frac{1}{2} k x^2 \]where \( k = 45.0 \) N/m and \( x = 0.280 \) m. Plugging in the values, we have:\[ SPE = \frac{1}{2} \times 45.0 \times (0.280)^2 = 1.764 \text{ J} \]

Step 2: Calculate Work Done by Friction (W_f)

The work done by friction over the distance the box moves while being in contact with the spring is calculated using:\[ W_f = \mu_k \cdot m \cdot g \cdot d \]where \( \mu_k = 0.300 \), \( m = 1.60 \) kg, \( g = 9.8 \) m/s² and \( d = 0.280 \) m. Thus:\[ W_f = 0.300 \times 1.60 \times 9.8 \times 0.280 = 1.31664 \text{ J} \]

Step 3: Apply Conservation of Energy to Find Speed

The kinetic energy (KE) of the box when it leaves the spring can be determined by the difference between the spring potential energy and the work done against friction:\[ KE = SPE - W_f = 1.764 - 1.31664 = 0.44736 \text{ J} \]Since KE is also given by \( \frac{1}{2} m v^2 \) where \( m = 1.60 \) kg, we can find velocity \( v \) by:\[ 0.44736 = \frac{1}{2} \times 1.60 \times v^2 \]\[ v^2 = \frac{0.44736 \times 2}{1.60} \]\[ v^2 = 0.558 \]\[ v = \sqrt{0.558} \approx 0.747 \text{ m/s} \]

Step 4: Determine Maximum Speed of Box

The maximum speed of the box occurs before friction has done any work on it. Initially, all the energy is stored as spring potential energy, which converts into kinetic energy. Thus, at maximum speed:\[ \frac{1}{2} m v^2_{\text{max}} = SPE \]Solving for \( v_{\text{max}} \):\[ \frac{1}{2} \times 1.60 \times v^2_{\text{max}} = 1.764 \]\[ v^2_{\text{max}} = \frac{1.764 \times 2}{1.60} \]\[ v^2_{\text{max}} = 2.205 \]\[ v_{\text{max}} = \sqrt{2.205} \approx 1.485 \text{ m/s} \]

Key Concepts

Spring Potential Energy

When a spring is compressed or stretched, it stores energy in the form of spring potential energy. The amount of energy stored depends on how much the spring is deformed and the spring's stiffness. The formula to find this energy is given by

• \( SPE = \frac{1}{2} k x^2 \)

where:

• \( k \) is the spring constant, indicating the stiffness of the spring.
• \( x \) is the distance the spring is compressed or stretched from its natural length.

In our exercise, a spring with a constant of 45.0 N/m is compressed by 0.280 m. By substituting these values into the formula, we find that the spring potential energy is 1.764 J. This energy is initially stored in the spring and is entirely mechanical energy that can potentially convert into other forms, like kinetic energy, when the spring is released.

Kinetic Friction

Kinetic friction is the force that opposes the movement of two surfaces sliding past each other. It acts whenever there's relative motion and depends on two main factors:

• The roughness of the surfaces in contact, represented by the coefficient of kinetic friction \( \mu_k \).
• The weight of the object being moved, often described in terms of normal force \( mg \) when moving horizontally.

For the box in our scenario, this kinetic friction force opposes its motion when released from the spring. It is calculated using the formula:

• \( F_{f} = \mu_k \cdot m \cdot g \)

This helps us to understand how much force needs to be overcome by the spring's energy to keep the box moving. With \( \mu_k = 0.300 \) and the mass being 1.60 kg, we are able to calculate the work done against this friction over a specified distance.

Work Done Against Friction

When an object moves and friction is present, some energy is used to overcome this force, referred to as 'work done against friction'. This work represents energy that is not available for useful purposes like speeding up the box. For a distance \( d \) over which friction acts, the work done is:

• \( W_{f} = F_{f} \cdot d \)

Incorporating friction into the energy calculations is crucial because it determines how much energy remains converted into kinetic energy and speed. Our example uses \( d = 0.280 \) m, and calculates the work done to overcome friction as 1.31664 J using the kinetic friction force. This value is used to adjust the spring potential energy to find the kinetic energy available after overcoming friction.

Maximum Speed of an Object

The maximum speed of an object, in this context, occurs when all the energy stored in the spring has been converted into kinetic energy, with no energy lost to friction. Once the spring potential energy starts converting into kinetic energy, the object accelerates until friction or other forces begin to reduce that energy. According to the conservation of energy, at maximum speed:

• \( \text{KE}_{\text{max}} = \text{SPE} \)

For the box scenario, the energy is entirely kinetic after the spring is released and before friction starts acting, leading to its peak speed of approximately 1.485 m/s. By calculating and understanding this speed, one can predict how fast the object will be moving when it first detaches from the spring.