Question

That same 0.005 kg is dropped off of the top of the Empire State building, which is about 443 meters tall. It hits the ground only going 5.28 m/s. How much energy, in J, was converted to heat as it fell? Round to two significant figures.

Answer 1

The amount of energy converted to heat as the object fell is approximately 21.71 J.

Step-by-step explanation:

The energy converted to heat as the object falls can be calculated using the principle of conservation of energy. The initial potential energy of the object at the top of the Empire State building is given by the formula PE = mgh, where m is the mass of the object (0.005 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the height (443 m). Plugging in the values, we can calculate the initial potential energy as 0.005 kg * 9.8 m/s^2 * 443 m = 21.78 J.

The final kinetic energy of the object before it hits the ground is given by the formula KE = 1/2 * mv^2, where m is the mass of the object (0.005 kg) and v is the velocity (5.28 m/s). Plugging in the values, we can calculate the final kinetic energy as 1/2 * 0.005 kg * (5.28 m/s)^2 = 0.069 J.

To determine the energy converted to heat, we subtract the final kinetic energy from the initial potential energy: 21.78 J - 0.069 J = 21.71 J. Therefore, the amount of energy converted to heat as the object fell is approximately 21.71 J.