Question
(a) Barney drops a 5.0 kg rock from a height of 92 om.How much energy does the rock have when it is 40 m from the ground? (b) How fast is the rock traveling at this point?
Solution
4.2
(295 Votes)
Fern
Professional · Tutor for 6 years
Answer
**(a) Calculating the rock's energy 40 m from the ground:**The rock initially possesses potential energy due to its height. As it falls, this potential energy is converted into kinetic energy. However, the *total* mechanical energy (potential + kinetic) remains constant, assuming we ignore air resistance. Therefore, we can calculate the potential energy at 40m and that will represent the total energy at that point.* **Formula for Potential Energy (PE):** PE = mgh * where m = mass (kg), g = acceleration due to gravity (9.8 m/s²), and h = height (m)* **Calculation:** PE = (5.0 kg) * (9.8 m/s²) * (40 m) PE = 1960 JoulesTherefore, the rock has 1960 Joules of energy when it is 40 m from the ground.**(b) Calculating the rock's speed at 40 m:**We can use the conservation of energy principle to find the rock's speed. The total energy at 40m (which we calculated above) is equal to the sum of the potential energy and kinetic energy at that point.1. **Calculate the potential energy at 40m (we already did this):** PE = 1960 Joules2. **Calculate the initial potential energy (at 92m):** PE_initial = (5.0 kg) * (9.8 m/s²) * (92 m) = 4508 Joules3. **Calculate the kinetic energy at 40m:** Since total energy is conserved, KE at 40m = Initial PE - PE at 40m KE = 4508 J - 1960 J = 2548 J4. **Use the kinetic energy formula to find the speed:** * Formula for Kinetic Energy (KE): KE = (1/2)mv² * where m = mass (kg) and v = velocity (m/s) * Rearranging the formula to solve for v: v = sqrt(2*KE/m) * Calculation: v = sqrt((2 * 2548 J) / 5.0 kg) v ≈ 31.9 m/sTherefore, the rock is traveling at approximately 31.9 m/s when it is 40 m from the ground.