Q1. A stick person slides down a hill inclined at 28° for 85 m. If the coefficient of friction is 0.090 and the mass is 60 kg, what is the speed at the bottom?

Background

Topic: Conservation of Energy with Friction

This question tests your understanding of energy conservation, specifically how gravitational potential energy is converted into kinetic energy and how friction (a non-conservative force) affects the final speed.

Key Terms and Formulas

• Gravitational Potential Energy:

• Kinetic Energy:

• Work done by friction:

• Kinetic friction force:

• Normal force on an incline:

• Height of incline:

Step-by-Step Guidance

1. Calculate the vertical height descended: .

2. Find the initial gravitational potential energy: .

3. Calculate the work done by friction: , where .

4. Set up the energy conservation equation: , where .

5. Rearrange the equation to solve for (the final speed at the bottom), but do not compute the final value yet.

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Q2. If the snow is level at the foot of the incline and has the same coefficient of friction, how far will the ski travel along the level?

Background

Topic: Work-Energy Principle with Friction

This question tests your ability to apply the work-energy theorem to motion on a horizontal surface with friction.

Key Terms and Formulas

• Initial kinetic energy: (from previous part)

• Work done by friction:

• Kinetic friction force: (on level ground)

• Work-Energy Principle: (since final speed is zero)

Step-by-Step Guidance

1. Use the final speed from the previous part as the initial speed on the level surface.

2. Set up the work-energy equation: .

3. Rearrange to solve for (distance traveled before stopping).

4. Plug in the known values for , , , and .

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Q3. The metabolic power for running at 15 km/hr is 1150 W. If a person runs for 30 min/day, how many calories do they use?

Background

Topic: Power, Energy, and Unit Conversion

This question tests your understanding of the relationship between power, energy, and time, as well as converting between joules and dietary calories.

Key Terms and Formulas

• Power:

• Energy:

• 1 dietary Calorie (kcal) = 4184 J

• Time conversion: 30 min = 1800 s

Step-by-Step Guidance

1. Convert 30 minutes to seconds.

2. Calculate the total energy used: .

3. Convert the energy from joules to kilocalories (Calories) using .

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Q4. A typical carrot has 125 kJ of energy. How many carrots do you need to eat to supply the energy used in the previous question?

Background

Topic: Energy Content and Unit Conversion

This question tests your ability to relate energy expenditure to food energy content.

Key Terms and Formulas

• Energy per carrot: J

• Total energy needed: from previous question

• Number of carrots:

Step-by-Step Guidance

1. Use the total energy calculated in the previous question.

2. Divide the total energy by the energy per carrot to find the number of carrots needed.

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Q5. How much waste energy does the runner generate?

Background

Topic: Efficiency and Energy Transformation

This question tests your understanding of how not all metabolic energy is converted to useful work; some is lost as waste heat.

Key Terms and Formulas

• Useful energy: energy used for running (from previous calculation)

• Total energy consumed: from food (carrots)

• Waste energy:

Step-by-Step Guidance

1. Determine the total energy intake (from carrots or Calories consumed).

2. Subtract the useful energy (mechanical work) from the total energy intake to find the waste energy.

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Q6. Assuming a typical person radiates at 186 W, how long would it take to radiate away the energy used in the run?

Background

Topic: Power, Energy, and Time

This question tests your ability to relate energy, power, and time, and to apply the formula to solve for time.

Key Terms and Formulas

• Power radiated: W

• Energy to be radiated: from previous calculation

• Time:

Step-by-Step Guidance

1. Use the total energy to be radiated (from previous answers).

2. Divide the energy by the power to find the time required.

3. Convert the time to appropriate units (e.g., minutes or hours).

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Q7. What is the rate of conduction of heat across only the 4 exterior walls of a 1600 ft² home (40 ft x 40 ft, 10 ft high, wall thickness 6.5 in, thermal conductivity 0.17 W/(m·K)), with inside temperature 70°F and outside 0°F?

Background

Topic: Heat Transfer by Conduction

This question tests your ability to calculate the rate of heat transfer through a wall using the thermal conductivity equation, and to convert units as needed.

Key Terms and Formulas

• Heat conduction rate:

• = thermal conductivity (W/(m·K))

• = total wall area (m²)

• = temperature difference (K or °C)

• = wall thickness (m)

Step-by-Step Guidance

1. Calculate the total area of the four walls: and convert to m².

2. Convert wall thickness from inches to meters.

3. Convert temperatures from °F to °C or K and find .

4. Plug all values into the formula to set up the calculation.

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Q8. If your house is radiating energy through the roof only (flat roof, emissivity 0.92), what is your home's rate of heat transfer by radiation through the roof?

Background

Topic: Heat Transfer by Radiation

This question tests your understanding of the Stefan-Boltzmann law for radiative heat transfer and how to apply it to a real-world scenario.

Key Terms and Formulas

• Stefan-Boltzmann law:

• = emissivity (0.92)

• = Stefan-Boltzmann constant ( W/m²·K⁴)

• = area of the roof (m²)

• , = inside and outside temperatures (in K)

Step-by-Step Guidance

1. Calculate the area of the roof: and convert to m².

2. Convert inside and outside temperatures from °F to K.

3. Plug all values into the Stefan-Boltzmann law to set up the calculation.

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