Question

3. A 3.0 × 10−3 kg copper penny drops a distance of 50.0 m to the ground. If 65 percent of the initial potential energy goes into increasing the internal energy of the penny, determine the magnitude of that increase.

Answer 1

The magnitude of the increase in internal energy of a 3.0 × 10⁻³ kg copper penny dropped from a height of 50.0 m, with 65 percent of its potential energy converted, is approximately 0.9555 J.

Step-by-step explanation:

The student is asking about the increase in the internal energy of the penny due to a fall. To determine the magnitude of this increase, we use the conservation of energy principle and the given percentage, which represents the portion of the initial potential energy that was converted into internal energy. The potential energy (PE) of the penny can be calculated using the formula PE = mgh, where m is the mass, g is the acceleration due to gravity, and h is the height.

For a penny of mass 3.0 × 10⁻³ kg, dropped from a height of 50.0 m, the potential energy at the start is PE = (3.0 × 10⁻³ kg)(9.8 m/s²)(50.0 m). Calculating this gives us PE = 1.47 J. Since 65 percent of the initial potential energy goes into increasing the internal energy of the penny, we multiply PE by 0.65, yielding an increase in internal energy of about 0.9555 J.

Answer 2

The magnitude of the increase is 0.9555J

Step-by-step explanation:

Potential Energy (PE) = mgh

mass (m) = 3×10^-3kg, acceleration due to gravity (g) = 9.8m/s^2, distance (h) = 50m

PE = 3×10^-3 × 9.8 × 50 = 1.47J

Increase in internal energy = 65% of 1.47J = 0.65 × 1.47J = 0.9555J