/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 33 A crate of fruit with mass 35.0 ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 17: Problem 33

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?

Short Answer

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

Step by step solution

01

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:\[ h = 8.0 \, \mathrm{m} \times \sin(36.9^{\circ}) \approx 4.8 \, \mathrm{m} \]\[ PE = 35.0 \, \mathrm{kg} \times 9.8 \, \mathrm{m/s}^2 \times 4.8 \, \mathrm{m} \approx 1646 \, \mathrm{J} \]
02

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:\[ KE = \frac{1}{2} \times 35.0 \, \mathrm{kg} \times (2.50 \, \mathrm{m/s})^2 = 109.375 \, \mathrm{J} \]
03

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:\[ W_{\text{friction}} = 1646 \, \mathrm{J} - 109.375 \, \mathrm{J} \approx 1536.625 \, \mathrm{J} \]
04

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 \(\Delta T\):\[ \Delta T = \frac{Q}{mc} = \frac{1536.625 \, \mathrm{J}}{35.0 \, \mathrm{kg} \times 3650 \, \mathrm{J/kg} \cdot \mathrm{K}} \approx 0.012 \, \mathrm{K} \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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/s虏), 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路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路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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The Sizes of Stars. The hot glowing surfaces of stars emit energy in the form of electromagnetic radiation. It is a good approximation to assume \(e=1\) for these surfaces. Find the radii of the following stars (assumed to be spherical): (a) Rigel, the bright blue star in the constellation Orion, which radiates energy at a rate of \(2.7 \times 10^{32} \mathrm{W}\) and has surface temperature \(11,000 \mathrm{K}\) ; (b) Procyon \(\mathrm{B}\) (visible only using a telescope), which radiates energy at a rate of \(2.1 \times 10^{23} \mathrm{W}\) and has surface temperature \(10,000 \mathrm{K}\) (c) Compare your answers to the radius of the earth, the radius of the sun, and the distance between the earth and the sun. (Rigel is an example of a supergiant star, and Procyon \(\mathrm{B}\) is an example of a white dwarf star.)

A Styrofoam bucket of negligible mass contains 1.75 \(\mathrm{kg}\) of water and 0.450 \(\mathrm{kg}\) of ice. More ice, from a refrigerator at \(-15.0^{\circ} \mathrm{C},\) is added to the mixture in the bucket, and when thermal equilibrium has been reached, the total mass of ice in the bucket is 0.868 \(\mathrm{kg} .\) Assuming no heat exchange with the surroundings, what mass of ice was added?

Spacecraft Reentry. A spacecraft made of aluminum circles the earth at a speed of 7700 \(\mathrm{m} / \mathrm{s} .\) (a) Find the ratio of its kinetic energy to the energy required to raise its temperature from \(0^{\circ} \mathrm{C}\) to \(600^{\circ} \mathrm{C}\) . (The melting point of aluminum is \(660^{\circ} \mathrm{C}\) . Assume a constant specific heat of 910 \(\mathrm{J} / \mathrm{kg} \cdot \mathrm{K} .\) (b) Discuss the bearing of your answer on the problem of the reentry of a manned space vehicle into the earth's atmosphere.

A technician measures the specific heat of an unidentified liquid by immersing an electrical resistor in it. Electrical energy is converted to heat transferred to the liquid for 120 s at a constant rate of 65.0 W. The mass of the liquid is \(0.780 \mathrm{kg},\) and its temperature increases from \(18.55^{\circ} \mathrm{C}\) to \(22.54^{\circ} \mathrm{C}\) . (a) Find the average specific heat of the liquid in this temperature range. Assume that negligible heat is transferred to the container that holds the liquid and that no heat is lost to the surroundings. (b) Suppose that in this experiment heat transfer from the liquid to the container or surroundings cannot be ignored. Is the result calculated in part (a) an overestimate or an underestimate of the average specific heat? Explain.

(a) Calculate the one temperature at which Fahrenheit and Celsius thermometers agree with each other. (b) Calculate the one temperature at which Fahrenheit and Kelvin thermometers agree with each other.
See all solutions

Recommended explanations on Physics Textbooks

Magnetism and Electromagnetic Induction

Read Explanation

Physical Quantities And Units

Read Explanation

Fluids

Read Explanation

Electrostatics

Read Explanation

Oscillations

Read Explanation

Thermodynamics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.