Problem 33 - University Physics with Modern Physics

Question

A crate of fruit with mass 35.0 \(\mathrm{kg}\) and specific heat 3650 \(\mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\) slides down a ramp inclined at \(36.9^{\circ}\) below the horizontal. The ramp is 8.00 \(\mathrm{m}\) long. (a) If the crate was at rest at the top of the incline and has a speed of 2.50 \(\mathrm{m} / \mathrm{s}\) at the bottom, how much work was done on the crate by friction? (b) If an amount of heat equal to the magnitude of the work done by friction goes into the crate of fruit and the fruit reaches a uniform final temperature, what is its temperature change?

Step-by-Step Solution

Answer

(a) Work done by friction is 1536.625 J. (b) Temperature change is 0.012 K.

Step 1: Calculate Potential Energy Lost

The potential energy lost by the crate as it slides down can be calculated using the formula \[ PE = mgh \]where \(m = 35.0 \, \mathrm{kg}\), \(g = 9.8 \, \mathrm{m/s}^2\), and the height \( h = L \sin \theta\) with \(L = 8.0 \, \mathrm{m}\) and \(\theta = 36.9^{\circ}\).Calculate:\[ h = 8.0 \, \mathrm{m} \times \sin(36.9^{\circ}) \approx 4.8 \, \mathrm{m} \]\[ PE = 35.0 \, \mathrm{kg} \times 9.8 \, \mathrm{m/s}^2 \times 4.8 \, \mathrm{m} \approx 1646 \, \mathrm{J} \]

Step 2: Calculate Kinetic Energy at the Bottom

The kinetic energy of the crate at the bottom of the ramp can be calculated using the formula \[ KE = \frac{1}{2} mv^2 \]where \(m = 35.0 \, \mathrm{kg}\) and \(v = 2.50 \, \mathrm{m/s}\).Calculate:\[ KE = \frac{1}{2} \times 35.0 \, \mathrm{kg} \times (2.50 \, \mathrm{m/s})^2 = 109.375 \, \mathrm{J} \]

Step 3: Calculate Work Done by Friction

To find the work done by friction, use the energy conservation equation:\[ PE_{\text{lost}} = KE + W_{\text{friction}} \]Rearrange to solve for the work done by friction:\[ W_{\text{friction}} = PE_{\text{lost}} - KE \]Substitute the previously calculated values:\[ W_{\text{friction}} = 1646 \, \mathrm{J} - 109.375 \, \mathrm{J} \approx 1536.625 \, \mathrm{J} \]

Step 4: Calculate Temperature Change

If the work done by friction is converted into heat, the temperature change can be found using \[ Q = mc\Delta T \]where \(Q = 1536.625 \, \mathrm{J}\), \(m = 35.0 \, \mathrm{kg}\), and \(c = 3650 \, \mathrm{J/kg} \cdot \mathrm{K}\).Rearrange to solve for \(\Delta T\):\[ \Delta T = \frac{Q}{mc} = \frac{1536.625 \, \mathrm{J}}{35.0 \, \mathrm{kg} \times 3650 \, \mathrm{J/kg} \cdot \mathrm{K}} \approx 0.012 \, \mathrm{K} \]

Key Concepts

Potential Energy

Potential energy is the stored energy of an object due to its position relative to a reference point, often related to the height above the ground. In the context of our example, potential energy is calculated when the crate is at the top of the ramp. The formula used for potential energy is: \( PE = mgh \). Here, \(m\) is the mass of the crate (35.0 kg), \(g\) stands for the acceleration due to gravity (9.8 m/s²), and \(h\) is the vertical height of the ramp.
The key to finding potential energy here was calculating the effective height from which the crate descends, which is determined by multiplying the length of the ramp by the sine of the ramp's angle: \( h = L \sin \theta \). With this data, calculating the potential energy as the crate moves became simplified and led to a resultant energy of approximately 1646 Joules.

Kinetic Energy

Kinetic energy is the energy possessed by an object due to its motion. When the crate reaches the bottom of the ramp, it has transformed some of its potential energy into kinetic energy. To compute kinetic energy, the formula \( KE = \frac{1}{2} mv^2 \) is used, where \(m\) represents mass (35.0 kg) and \(v\) is the velocity of the crate (2.50 m/s) at the bottom.
The method involves squaring the velocity, multiplying by the mass, and then dividing by two. It captures the essence of how speed and mass contribute to the energy, resulting in a lower energy value here of 109.375 Joules, which indicates that not all potential energy converted into kinetic energy due to friction.

Specific Heat Capacity

Specific heat capacity refers to the amount of heat energy required to raise the temperature of a unit mass of a substance by one degree Kelvin (or Celsius). In this exercise, the specific heat capacity of the crate's material is given as 3650 J/(kg·K). This value indicates the material's ability to absorb heat.
Knowing the specific heat capacity is crucial for determining how much temperature change will occur when the crate absorbs energy, like from frictional heat. The specific heat capacity is intrinsic to the material the crate is made of, influencing how it reacts to energy input and temperature variance.

Temperature Change

Temperature change is determined by how much heat energy the crate absorbs or releases, considering its mass and specific heat capacity. In this scenario, the work done by friction as the crate slides down is transformed into heat energy, causing a temperature change.
The formula \( Q = mc\Delta T \) relates heat (\(Q\)) to temperature change \(\Delta T\), where \(m\) is mass and \(c\) specific heat capacity. Rearranging gives \( \Delta T = \frac{Q}{mc} \), and calculating it with \( Q = 1536.625 \) J, \( m = 35.0 \) kg, and \( c = 3650 \) J/(kg·K), results in a minimal temperature increase around 0.012 K.
This negligible change signifies how the material's properties impact heat absorption, making it crucial in understanding energy conservation's thermal aspects.