Menu div.question,div.steps {font-size: 16px; padding: 10px;line-height: 20px; margin: 20px;} div.question{border: 1px solid #ccc;} div#solution .button{font-size: 20px; margin: 20px; width: 200px; padding: 10px; text-align: center;} div.hint,div.answer,div.steps{display: none;} div.pre_next{font-size: 30px; width:90%; margin:auto;} div.pre_next a.nav_pre{float:left;} div.pre_next a.nav_next{float:right;} div.title div.notes{ font-family:sans-serif, Arial; font-size: 18px; } div.speak{display:none;} i.speak{float: right;margin-top:-10px;cursor:pointer;} 3. A 1000-kg elevator is moving from 10\ m above the ground to 20\ m above the ground. (a) What is the initial potential energy of the elevator. (b) What is the final potential energy of the elevator. (c) Calculate the work done by the lifting force.3. A 1000-kg elevator is moving from 10\ m above the ground to 20\ m above the ground. (a) What is the initial potential energy of the elevator. (b) What is the final potential energy of the elevator. (c) Calculate the work done by the lifting force. Hint PE = mgh W = ∆PE Answer (a) PE_i = 1.0 times 10^5\ J (b) PE_f = 2.0 times 10^5\ J (c) W = 1.0 times 10^5\ J Show Steps Given m = 1000\ kg h_i = 10\ m h_f = 20\ m Equations: PE = mgh W = ∆PE Solution (a) PE_i = mgh_i PE_i = (1000\ kg)(10\ m//s^2)(10\ m) PE_i = 1.0 times 10^5\ J (b) PE_f = mgh_f PE_f = (1000\ kg)(10\ m//s^2)(20\ m) PE_f = 2.0 times 10^5\ J (c) W = ∆PE = PE_f - PE_i W = ( 2.0 times 10^5\ J)-( 1.0 times 10^5\ J) W = 1.0 times 10^5\ J