Question

kinetic energy is the type energy that depends on movement and is based on an objects mass and velocity. an objects potential energy is its stored energy due to its position. for a closed system, the sum of an objects kinetic and potential energy remains constant. potential energy: p.e. = weight x height, p.e. = w x h kinetic energy: ke = ½ mass x velocity², k.e. = ½ mv² the metrics units used are: energy = joules weight = newtons (i.e., kg x 9.8 n/kg) height = meters mass = kilograms velocity = m/s provide the answers to the questions below. 1. consider a rock weighing 100 newtons that is at the top of a 200 meter high hill. what is its potential energy? answer: 2. consider a a bicycle with a mass of 20 kg traveling at a velocity of 4 m/s. what is its kinetic energy? answer: 3. consider a flower pot weighing 4 newtons which is sitting on a windowsill 30 meters from the ground. how many joules of kinetic and potential energy does it have? answer: 4. if the flower pot in problem 3 falls off the windowsill and falls 20 meters downwards (i.e., it is 10 meters from hitting the ground), what is its potential energy at that point? answer: 5. using your answer in question 4, what kinetic energy does the flower pot have when it has fallen 20 meters and is 10 meters from the ground? answer: 6. using your answer from question 5, what speed is the flower pot traveling when it has fallen 20 meters and is 10 meters from the ground? answer:

Explanation:

Step1: Calculate potential energy for question 1

Use the formula \(P.E. = w\times h\). Given \(w = 100\space N\) and \(h=200\space m\), then \(P.E.=100\times200\)

Step2: Calculate kinetic energy for question 2

Use the formula \(K.E.=\frac{1}{2}mv^{2}\). Given \(m = 20\space kg\) and \(v = 4\space m/s\), then \(K.E.=\frac{1}{2}\times20\times4^{2}\)

Step3: Analyze question 3

A flower - pot sitting on a windowsill has \(v = 0\space m/s\), so \(K.E. = 0\space J\). Use \(P.E.=w\times h\) with \(w = 4\space N\) and \(h = 30\space m\), so \(P.E.=4\times30\)

Step4: Calculate potential energy for question 4

Use \(P.E.=w\times h\). Now \(h = 10\space m\) and \(w = 4\space N\), so \(P.E.=4\times10\)

Step5: Use energy conservation for question 5

The total initial energy \(E=P.E._{initial}=4\times30=120\space J\). At \(h = 10\space m\), \(P.E.=40\space J\). Then \(K.E.=E - P.E.\), so \(K.E.=120 - 40\)

Step6: Calculate velocity for question 6

Use \(K.E.=\frac{1}{2}mv^{2}\). First, find \(m=\frac{w}{g}=\frac{4}{9.8}\space kg\). Given \(K.E. = 80\space J\), then \(80=\frac{1}{2}\times\frac{4}{9.8}\times v^{2}\), solve for \(v\)

Answer:

1. \(20000\space J\)
2. \(160\space J\)
3. \(K.E. = 0\space J\), \(P.E.=120\space J\)
4. \(40\space J\)
5. \(80\space J\)
6. \(v=\sqrt{\frac{2\times80\times9.8}{4}}\approx19.8\space m/s\)