Problem 65 - University Physics with Modern Physics

Question

In a truck loading station at a post office, a small 0.200 \(\mathrm{kg}\) package is released from rest at point \(A\) on a track that is one-quarter of a circle with radius 1.60 \(\mathrm{m}(\mathrm{Fig} .7 .39) .\) The size of the package is much less than \(1.60 \mathrm{m},\) so the package can be treated as a particle. It slides down the track and reaches point \(B\) with a speed of 4.80 \(\mathrm{m} / \mathrm{s}\) . From point \(B,\) it slides on a level surface a distance of 3.00 \(\mathrm{m}\) to point \(C,\) where it comes to rest (a) What is the coefficient of kinetic friction on the horizontal surface? (b) How much work is done on the package by friction as it slides down the circular are from \(A\) to \(B ?\)

Step-by-Step Solution

Answer

(a) The coefficient of kinetic friction \( \mu_k \) is 0.392. (b) The work done by friction from A to B is \(-1.57 \text{J}\).

Step 1: Define the Problem Variables

We have a package with mass \( m = 0.200 \, \text{kg} \), a radius of the circular arc \( r = 1.60 \, \text{m} \), initial speed \( v_A = 0 \, \text{m/s} \), final speed at point B \( v_B = 4.80 \, \text{m/s} \), and sliding distance on a horizontal surface \( d = 3.00 \, \text{m} \). We need to find the coefficient of kinetic friction \( \mu_k \) for the surface from B to C and the work done by friction from A to B.

Step 2: Use Conservation of Energy on the Circular Track

The package starts with gravitational potential energy at point A and ends with kinetic energy at point B. Use the conservation of mechanical energy equation: \[ mgh_A = \frac{1}{2}mv_B^2 + W_{friction} \]Here, \( h_A = r \cdot (1 - \cos(\theta)) \) is the height drop, where \( \theta = 90^\circ \rightarrow \cos(\theta) = 0 \), and \( W_{friction} \) is the work done by friction, which will be calculated later.

Step 3: Calculate Potential Energy and Kinetic Energy

The height at point A is \( h_A = 1.60 \, \text{m} \). The potential energy at A is:\[ PE_A = mgh_A = 0.200 \times 9.8 \times 1.60 \]The kinetic energy at B is:\[ KE_B = \frac{1}{2}mv_B^2 = \frac{1}{2} \times 0.200 \times (4.80)^2 \]

Step 4: Determine Work Done by Friction from B to C

On the horizontal surface, the work done by kinetic friction brings the package to rest: \[ W_{friction, BC} = \mu_k mgd \]We know that the package comes to rest after sliding 3.00 m, so kinetic energy at point B is entirely dissipated.

Step 5: Solve for \(\mu_k\) on the Horizontal Surface

Using the work-energy principle, equate the kinetic energy at B to the work done by friction from B to C:\[ \frac{1}{2}mv_B^2 = \mu_k \cdot mg \cdot 3.00 \]Solving for \( \mu_k \):\[ \mu_k = \frac{\frac{1}{2}mv_B^2}{mg \cdot 3.00} \]

Step 6: Calculate Work Done by Friction on the Circular Arc

Return to the conservation of energy equation from Step 2 and solve for \( W_{friction} \) on the arc:\[ mgh_A - \frac{1}{2}mv_B^2 = W_{friction, AB} \]Calculate \( W_{friction, AB} \) substituting the values of \( m, g, h_A, \) and \( v_B \).

Key Concepts

Kinetic Friction

Kinetic friction is an essential concept when studying objects moving across surfaces, including the scenario where our package moves from point B to C. This type of friction occurs when two surfaces slide against each other, creating a resistance to the motion. The magnitude of this force is typically calculated using the coefficient of kinetic friction (\( \mu_k \)) and the normal force (\( F_n \)) acting on the object.
To put it simply, the kinetic friction force \( F_{friction} \) can be expressed using:

• \( F_{friction} = \mu_k \times F_n \)

For the package sliding on the horizontal surface, the normal force is simply the weight of the package, \( mg \), given that there are no other vertical forces acting. Hence, the kinetic friction force becomes \( F_{friction} = \mu_k \times mg \). As the package moves from point B to C, this frictional force does work on it, which can be calculated through the formula for work done by a force:

• \( W_{friction, BC} = \mu_k \times mg \times d \)

where \( d \) is the distance traveled. Through understanding and calculating these values, we can see how kinetic friction affects the motion of objects.

Mechanical Energy

Mechanical energy is the sum of potential energy and kinetic energy of an object. In many physics problems, particularly those involving motion like our package's journey, conservation of mechanical energy is a critical principle. It states that in an isolated system (with no external work done), mechanical energy remains constant.
For our package, the initial mechanical energy at point A includes only gravitational potential energy because it starts at rest, which is given by:

• \( PE_A = mgh_A \)

As the package slides down to point B, this potential energy transforms into kinetic energy, depicted by:

• \( KE_B = \frac{1}{2}mv_B^2 \)

However, not all of the initial potential energy turns into kinetic energy due to the work done by friction as it slides. The conservation of mechanical energy can be represented by the equation:

• \( mgh_A = \frac{1}{2}mv_B^2 + W_{friction} \)

This equation helps us understand how potential energy, kinetic energy, and work by friction interact and balance within this system.

Work-Energy Principle

The work-energy principle is a foundational concept in physics that relates the work done on an object to the change in its kinetic energy. This principle underpins many calculations involving energy transformations and is integral to both parts of our package problem, from the circular arc to the horizontal slide.
The work-energy principle states:

• The work done by all forces acting on an object equals the change in the object's kinetic energy.

Initially, as the package moves from A to B, gravitational force does positive work, converting potential energy into kinetic energy while friction does negative work, absorbing some of this energy. This can be expressed as:

• \( mgh_A - W_{friction, AB} = \frac{1}{2}mv_B^2 \)

On the horizontal path from B to C, the work-energy principle tells us that the work done by kinetic friction equals the loss of kinetic energy, bringing the package to rest:

• \( \frac{1}{2}mv_B^2 = W_{friction, BC} \)

By applying the work-energy principle, we can determine how forces like friction affect the energy and motion of objects.