Question

Here's an enhanced version of your questions:

In a lab investigation, one group of students (group A) measures the speed of a 0.1-kilogram car at 2.5 m/sec at the bottom of a hill. Another group of students (group B) measures the speed of the car at 3 m/sec at the bottom of the hill. The car’s starting position at the top of the hill is one meter high.
a. What is the potential energy of the car at the beginning of the experiment before its speed is measured?

b. Calculate the kinetic energy of the car for group A using the speed (2.5 m/sec) and mass values above.

c. Calculate the kinetic energy of the car for group B using the speed (3.0 m/sec) and mass values above.

d. At the bottom of a hill, the kinetic energy of the cars should be equal to the potential energy of the car at the top of the hill. Are the kinetic energy values for groups A and B equal to, less than, or greater than the potential energy value?

e. The energy of an object can be converted to heat due to the friction of the car on the hill. The difference between the potential energy of the car and its kinetic energy at the bottom of the hill equals the energy lost due to friction. How much energy is lost due to heat for group A’s car? How much for group B’s?

Answer 1

The potential energy of the car at the start is 0.98 joules. Group A's car has a kinetic energy of 0.3125 joules and group B's car has 0.45 joules. The kinetic energy for both groups is less than the potential energy, indicating energy lost to heat.

Step-by-step explanation:

The question deals with the conversion of gravitational potential energy into kinetic energy as a car travels down a hill. To answer the student's questions:

1. Potential energy at the start is calculated using the formula PE = mgh. Since the mass m is 0.1 kg, gravity g is 9.8 m/s2, and the height h is 1 m, the potential energy PE is 0.1 kg * 9.8 m/s2 * 1 m = 0.98 joules.
2. The kinetic energy for group A is KE = 0.5 * m * v2. Substituting m with 0.1 kg and v with 2.5 m/sec gives KE = 0.5 * 0.1 kg * (2.5 m/sec)2 = 0.3125 joules.
3. For group B, with v as 3 m/sec, KE = 0.5 * 0.1 kg * (3 m/sec)2 = 0.45 joules.
4. The kinetic energy values for both groups A and B are less than the potential energy at the top of the hill.
5. The energy lost to heat for group A is 0.98 joules (potential energy) - 0.3125 joules (kinetic energy) = 0.6675 joules. For group B, it's 0.98 joules - 0.45 joules = 0.53 joules.