6. A 2 kg mass is dropped to the ground through a conducting loop. The mass starts 3 meters off the ground with zero velocity, and it lands with a velocity 7 meters/second. You may take g to be 9.8 m/s^2 and ignore all other forces than gravity and electricity + magnetism. How much energy did the conducting loop dissipate (release) as heat and how do you know?
Energy transfer
The flow of energy from one region to another region is referred to as energy transfer. Since energy is quantitative; it must be transferred to a body or a material to work or to heat the system.
Molar Specific Heat
Heat capacity is the amount of heat energy absorbed or released by a chemical substance per the change in temperature of that substance. The change in heat is also called enthalpy. The SI unit of heat capacity is Joules per Kelvin, which is (J K-1)
Thermal Properties of Matter
Thermal energy is described as one of the form of heat energy which flows from one body of higher temperature to the other with the lower temperature when these two bodies are placed in contact to each other. Heat is described as the form of energy which is transferred between the two systems or in between the systems and their surrounding by the virtue of difference in temperature. Calorimetry is that branch of science which helps in measuring the changes which are taking place in the heat energy of a given body.
![**Problem 6:**
A 2 kg mass is dropped to the ground through a conducting loop. The mass starts 3 meters off the ground with zero velocity, and it lands with a velocity of 7 meters/second. You may take \( g \) to be \( 9.8 \, \text{m/s}^2 \) and ignore all other forces than gravity and electricity + magnetism. How much energy did the conducting loop dissipate (release) as heat and how do you know?
**Solution Approach:**
1. **Initial Energy Calculation:**
- Calculate the initial potential energy (\( PE_i \)) using \( mgh \), where \( m \) is mass, \( g \) is acceleration due to gravity, and \( h \) is height.
- For this problem:
\[ PE_i = mgh = 2 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 3 \, \text{m} \]
2. **Final Energy Calculation:**
- Calculate the final kinetic energy (\( KE_f \)) using \( \frac{1}{2} mv^2 \), where \( m \) is mass and \( v \) is velocity.
- For this problem:
\[ KE_f = \frac{1}{2} \times 2 \, \text{kg} \times (7 \, \text{m/s})^2 \]
3. **Energy Dissipation:**
- The energy dissipated as heat by the conducting loop is the difference between the initial energy and the final kinetic energy of the mass.
- Thus, the energy dissipated:
\[ \text{Energy Dissipated} = PE_i - KE_f \]
Through these calculations, you'll determine how much energy the conducting loop dissipates as heat.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F60af46cc-0d0a-43f7-b2e0-84e1176374a1%2Fa532aea7-f1c6-4068-a21f-55e368b5b1ec%2F3sx1zca7_processed.jpeg&w=3840&q=75)
Given:
mass, m = 2 kg
initial height of mass, h = 3 m
initial velocity, u = 0
final velocity, v = 7 m/s
final height = 0
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