Problem 39 - 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 \(36.9^{\circ} \mathrm{C}\) 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 by friction is -1534.6 J. (b) Temperature change is 0.012 K.

Step 1: Calculate the Potential Energy at the Top

The potential energy (PE) at the top of the ramp is determined by the formula: \[ \text{PE} = m \cdot g \cdot h \]where \( m = 35.0 \, \text{kg} \) is the mass, \( g = 9.81 \, \text{m/s}^2 \) is the acceleration due to gravity, and \( h \) is the height. To find \( h \), use trigonometry: \[ h = 8.00 \, \text{m} \cdot \sin(36.9^{\circ}) \approx 4.80 \, \text{m} \]Now, calculate the potential energy:\[ \text{PE} = 35.0 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 \cdot 4.80 \, \text{m} \approx 1644 \, \text{J} \]

Step 2: Calculate the Kinetic Energy at the Bottom

The kinetic energy (KE) at the bottom of the ramp is calculated using the formula: \[ \text{KE} = \frac{1}{2} m v^2 \]where \( v = 2.50 \, \text{m/s} \) is the velocity. Substituting the known values, we get:\[ \text{KE} = \frac{1}{2} \cdot 35.0 \, \text{kg} \cdot (2.50 \, \text{m/s})^2 = 109.375 \, \text{J} \]

Step 3: Calculate the Work Done by Friction

The work done by friction \( W_f \) can be found using the work-energy principle, which states that the work done by non-conservative forces is equal to the change in mechanical energy (potential plus kinetic energy). \[ W_f = \text{KE} - \text{PE} \]Substitute the values:\[ W_f = 109.375 \, \text{J} - 1644 \, \text{J} = -1534.625 \, \text{J} \]The negative sign indicates that the frictional force did negative work (opposing the motion).

Step 4: Calculate the Temperature Change

The work done by friction is entirely converted into heat energy absorbed by the crate, causing a temperature change. The formula relating heat energy to temperature change \( \Delta T \) is:\[ Q = m \cdot c \cdot \Delta T \]where \( Q = 1534.625 \, \text{J} \) (positive amount of heat added), \( c = 3650 \, \text{J/kg} \cdot \text{K} \) is the specific heat capacity. Solving for \( \Delta T \):\[ \Delta T = \frac{Q}{m \cdot c} = \frac{1534.625 \, \text{J}}{35.0 \, \text{kg} \times 3650 \, \text{J/kg} \cdot \text{K}} \approx 0.012 \, \text{K} \]

Key Concepts

Friction

Friction is the force that resists the relative motion of objects sliding against each other. In this exercise, the crate experiences friction while it slides down the ramp, opposing its motion and transforming some of its mechanical energy into heat energy. Due to friction, the kinetic energy gained by the crate is less than the potential energy lost. This is why the work done by friction is crucial to calculate, as it equals the difference between the initial potential energy and the final kinetic energy of the crate.

The work done by friction, noted here as a negative value, signifies that the frictional force is opposing the crate's motion down the ramp. This is important to realize because, in physics, negative work implies that the force is acting in the opposite direction of the displacement of the object.

• Friction converts mechanical energy into heat, affecting the temperature of the materials involved.
• It plays a critical role in energy loss in sliding motions.
• In this exercise, the frictional work done is linked to how much the temperature of the crate will change.

Specific Heat Capacity

Specific heat capacity is a property that defines how much heat energy is required to raise the temperature of a unit mass of a substance by one degree Celsius (or one Kelvin). In this problem, the specific heat capacity of the crate’s contents is crucial to determine the temperature change due to the work done by friction.

Here, specific heat capacity (denoted as \( c \)) of the fruit mass is given as \( 3650 \, \text{J/kg} \cdot \text{K} \). This means it takes 3650 Joules of energy to raise the temperature of 1 kilogram of the crate's fruit by 1 Kelvin.

• This property depends on the material, influencing how quickly it heats up or cools down.
• Higher specific heat capacity indicates that the material can absorb more heat without undergoing a significant temperature change.
• Crucial for determining how the thermal energy impacts the temperature of an object.

Temperature Change

Temperature change refers to the variation in heat level of a material as it absorbs or releases thermal energy. In energy-related problems, calculating the temperature change is essential to understand how energy interactions affect the object.

In this exercise, the temperature change \( \Delta T \) of the crate can be calculated using the formula:
\[ Q = m \cdot c \cdot \Delta T \]
where \( Q \) is the heat energy transferred due to friction, \( m \) is the mass of the crate, and \( c \) is the specific heat capacity. The small temperature increase here (approximately 0.012 K) demonstrates how much the energy transfer from friction affects the crate’s temperature.

• Even a small amount of energy (in this context) can lead to temperature changes due to the specific heat properties.
• This connects the energy loss due to friction with the thermal dynamics of the object.
• The heat generated can affect material properties, efficiency, and safety in practical applications.