Don (82.8 kg) is going bungee jumping. The bridge he will jump off of is 54.0 m above the ground. After Don jumps, he comes to a spot 10.3 m above the ground before the bungee pulls him back.How much elastic energy is in the bungee cord when Don is 10.3 meters above the ground? If 6,200 J of the elastic energy (use your number from above) is converted to internal energy, to what height will Don return when the bungee bounces him back up? If Don drops his phone (0.18 kg) when he is 28.4 m in the air, what speed will it reach before it hits the ground?
Question
Don (82.8 kg) is going bungee jumping. The bridge he will jump off of is 54.0 m above the ground. After Don jumps, he comes to a spot 10.3 m above the ground before the bungee pulls him back.How much elastic energy is in the bungee cord when Don is 10.3 meters above the ground? If 6,200 J of the elastic energy (use your number from above) is converted to internal energy, to what height will Don return when the bungee bounces him back up? If Don drops his phone (0.18 kg) when he is 28.4 m in the air, what speed will it reach before it hits the ground?
Solution
To solve this problem, we need to use the principles of conservation of energy and the equations for gravitational potential energy and kinetic energy.
- First, we need to calculate the total energy when Don is at the highest point of his jump. This is the sum of his gravitational potential energy and the elastic energy in the bungee cord.
Gravitational potential energy (PE) is given by the formula PE = mgh, where m is mass, g is the acceleration due to gravity (9.8 m/s^2), and h is height.
PE = 82.8 kg * 9.8 m/s^2 * (54.0 m - 10.3 m) = 35,294.64 J
This is the elastic energy in the bungee cord when Don is 10.3 meters above the ground.
- If 6,200 J of the elastic energy is converted to internal energy, the remaining energy will be converted to gravitational potential energy when Don bounces back up.
Remaining energy = 35,294.64 J - 6,200 J = 29,094.64 J
We can set this equal to the gravitational potential energy at the height Don will return to and solve for h:
29,094.64 J = 82.8 kg * 9.8 m/s^2 * h h = 29,094.64 J / (82.8 kg * 9.8 m/s^2) = 36 m
So, Don will return to a height of 36 meters.
- If Don drops his phone when he is 28.4 m in the air, we can calculate the speed it will reach before it hits the ground using the equation for kinetic energy, KE = 1/2 mv^2, and setting it equal to the gravitational potential energy at the height the phone is dropped:
PE = mgh = 0.18 kg * 9.8 m/s^2 * 28.4 m = 50.05 J
Setting this equal to the kinetic energy and solving for v gives:
50.05 J = 1/2 * 0.18 kg * v^2 v^2 = 50.05 J / (1/2 * 0.18 kg) = 556.11 v = sqrt(556.11) = 23.6 m/s
So, the phone will reach a speed of 23.6 m/s before it hits the ground.
Similar Questions
If 6,200 J of the elastic energy (use your number from above) is converted to internal energy, to what height will Don return when the bungee bounces him back up?
A bungee jumper has a mass of 67 kg.The total height of the jump towards Earth is 210 m.Calculate the speed of the bungee jumper at the bottom of the jump just before the bungee cord stretches and catches them.Gravitational field strength = 9.8 N/kgIgnore the effects of air resistance.Give your answer to three significant figures.
How much potential energy does Tim, with mass 64.0 kg, gain when he climbs a gymnasium rope a distance of 3.3 m?
In an effort to impress Mr. Hass during PE class, Avery quickly elevates his 62.6-kg body up the rope to a height of 5.69 meters above his starting point. What is Avery's change in potential energy?
Chris jumps off a bridge with a bungee cord (a heavy stretchable cord) tied around his ankle . He falls for 15 m before the bungee cord begins to stretch. Chris's mass is 75 kg and we assume the cord obeys Hooke's law, F=−kx , with k = 60 N/m .
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.