a lump of lead with mass 0.50kg is dropped from a height of 20m onto a hard surface. it does not rebound but remains there at rest for a long period of time. what are heating asystem ? , mechanical work and .u
student
-100
To determine the heating system, mechanical work, and internal energy (.U) in this scenario, let's break it down step by step.
1. Heating System: When the lump of lead hits the hard surface, some of its kinetic energy is converted into thermal energy, causing it to heat up. We can calculate the heating system using the principle of conservation of energy.
The potential energy (PE) of the lump at the initial height (20m) is given by PE = mgh, where m is the mass, g is the acceleration due to gravity (9.8 m/s^2), and h is the height.
PE = (0.50 kg) * (9.8 m/s^2) * (20 m) = 98 J
Since there is no rebound and the lump remains at rest, all its initial potential energy is converted into thermal energy. Therefore, the heating system in this case is 98 J.
2. Mechanical Work: The lump of lead does not perform any mechanical work on the hard surface as there is no displacement in the direction of the applied force. Mechanical work is defined as the product of force and displacement in the direction of the force. Since the lump remains at rest, the displacement is zero, and thus the mechanical work is also zero.
3. Internal Energy (.U): The internal energy is a state function that describes the total energy of a system. In this case, as the lump of lead comes to rest and remains there, there is no change in internal energy. Therefore, the internal energy (.U) is zero.
To summarize:
- Heating System: 98 J
- Mechanical Work: 0 J
- Internal Energy (.U): 0 J