Problem 70 - University Physics with Modern Physics

Question

Engineers are designing a system by which a falling mass \(m\) imparts kinetic energy to a rotating uniform drum to which it is attached by thin, very light wire wrapped around the rim of the drum (Fig. 9.34\()\) . There is no appreciable friction in the axle of the drum, and everything starts from rest. This system is being tested on earth, but it is to be used on Mars, where the acceleration due to gravity is 3.71 \(\mathrm{m} / \mathrm{s}^{2} .\) In the earth tests, when \(m\) is set to 15.0 \(\mathrm{kg}\) and allowed to fall through \(5.00 \mathrm{m},\) it gives 250.0 \(\mathrm{J}\) of kinetic energy to the drum. (a) If the system is operated on Mars, through what distance would the \(15.0-0\) mass have to fall to give the same amount of kinetic energy to the drum? (b) How fast would the 15.0 \(\mathrm{kg}\) mass be moving on Mars just as the drum gained 250.0 \(\mathrm{J}\) of kinetic energy?

Step-by-Step Solution

Answer

(a) 13.2 m; (b) 10.0 m/s

Step 1: Understand the Concept of Energy Transfer

Before solving the problem, we need to understand that when the mass falls, potential energy is converted into kinetic energy of the drum. On Earth, a 15.0 kg mass falling through a height of 5.0 m provides 250.0 J of kinetic energy to the drum. We will use this understanding to solve the problem for Mars.

Step 2: Calculate Potential Energy Change on Earth

The potential energy change as the mass falls on Earth is given by the formula \( \Delta PE = mgh \). With \( m = 15.0 kg \), \( g = 9.81 \, m/s^2 \), and \( h = 5.0 \, m \), calculate the potential energy change: \[ \Delta PE = 15.0 \times 9.81 \times 5.0 = 735 \, J \].

Step 3: Determine Frictionless Energy Loss on Earth

Since only 250.0 J is given to the drum as kinetic energy, find the energy lost or not transferred: \[ \text{Energy not transferred} = 735 \, J - 250 \, J = 485 \, J \].

Step 4: Set Up Energy Balance Equation for Mars

On Mars, we want the drum to gain the same 250.0 J. Use the equation for potential energy, now on Mars: \( \Delta PE_{mars} = mgh_{mars} \). Let \( h_{mars} \) be the necessary height.

Step 5: Calculate Potential Energy Required on Mars

On Mars, use the formula \( mgh_{mars} - 485 \, J = 250 \, J \) since, similar to Earth, there is an expected loss or missing transfer of 485 J. Substitute the known values: \( g_{mars} = 3.71 \, m/s^2 \). Rearrange and solve for \(h_{mars}\): \[ 15.0 \cdot 3.71 \cdot h_{mars} - 485 = 250 \]. Solving, \[ 55.65 \cdot h_{mars} - 485 = 250 \]. \[ 55.65 \cdot h_{mars} = 735 \]. \[ h_{mars} = \frac{735}{55.65} \approx 13.2 \, m \].

Step 6: Determine Velocity of the Mass on Mars

Use kinetic energy formula to find final velocity of mass: \[ \frac{1}{2} mv^2 = KE_{mass} \]. Substituting, \( KE_{mass} = \Delta PE - 250 \, J \) = \( 15 \times 3.71 \times 13.2 - 250 \): \[ v = \sqrt{\frac{2 \times (15 \times 3.71 \times 13.2 - 250)}{15}} \]. \[ v \approx 10.0 \, m/s \].

Key Concepts

Potential Energy

Potential energy is the stored energy of an object due to its position in a gravitational field. For instance, when you lift an object off the ground, it gains potential energy. This is because it has the potential to do work due to its elevated position. The formula to calculate the potential energy when an object is lifted to a height is given by \( PE = mgh \), where:

• \( m \) is the mass of the object,
• \( g \) is the acceleration due to gravity,
• \( h \) is the height above the reference point.

On Earth, the gravitational acceleration \( g \) is approximately \( 9.81 \, m/s^2 \). So, if you have a 15 kg object and you lift it 5 meters, its potential energy would be \( 15 \, kg \times 9.81 \, m/s^2 \times 5 \, m = 735 \, J \).
On Mars, with a lower gravity of \( 3.71 \, m/s^2 \), the same mass lifted the same height would have less potential energy. This difference is crucial for energy transfer systems designed to work on different planets.

Energy Transfer

Energy transfer is the process by which energy changes form and moves from one object to another. In the problem, the falling mass allows for potential energy to transfer into kinetic energy as it descends. As the mass falls, its potential energy decreases while the kinetic energy of the drum increases, which is an example of energy conversion.
In ideal conditions, all potential energy converts into kinetic energy, but in real-world scenarios, some energy is always lost to inefficiencies.
In the Earth test, the system loses 485 J of potential energy which doesn't convert into kinetic energy for the drum. Understanding how and why this loss occurs helps improve energy transfer systems. On Mars, it's crucial to calculate how much height is necessary to achieve the same kinetic output despite lower energy transfer efficiency.

Gravitational Acceleration

Gravitational acceleration \( g \) is the rate of increase of velocity that an object experiences when it falls freely under gravity. It varies based on the celestial body you're on. For Earth, \( g \) is about \( 9.81 \, m/s^2 \), whereas on Mars, it's significantly lower at \( 3.71 \, m/s^2 \).
This difference affects potential energy calculations directly. An object on Mars requires a greater height to achieve the same potential energy as it would on Earth, given the lower gravitational pull.
In engineering systems, understanding gravitational differences is crucial when designing equipment meant for other planets. Calculations involving gravity must be adjusted to ensure that devices operate correctly in their specific environments.

Physics Problem Solving

Physics problem solving entails breaking down a problem into manageable parts, applying relevant laws consistently, and computing accurate results. It's crucial to understand each element of a problem to solve it efficiently. In the provided exercise, the goal was to find necessary height for potential energy conversion on Mars and the velocity of the mass.
Key steps for solving such a problem include:

• Understanding the context and setting up the problem,
• Applying conservation laws, like energy conservation,
• Using appropriate formulas and plugging in values,
• Understanding how each variable interacts, like mass, gravity, and height.

By using these steps, you ensure that the result is based on a logical approach and adheres to the laws of physics, helping to design systems that perform well under specified conditions.