Question

< Question 1 of 25 > Label the diagram with the correct values for kinetic energy (KE) and potential energy (PE) along the path of the roller coaster. Assume that the roller coaster starts from rest. Answer Bank KE: 2250.0 J PE: 2250.0 J KE: OJ PE: 4500.0 J KE: 3375.0J PE: 1125.0 J KE: 4500.0J PE: OJ KE: 1125.0 J PE: 3375.0J © 2011-2020 Sapling Learning, Inc. about us careers privacy policy terms of use contact us help

Answer 1

Potential Energy – 94 is the energy held by an object because of its (P.6)” position relative to the plane of reference, TP.E = mgh/ m= mass of object - g - acceleration due to gravity U=9.8 m/s2 h = height of object with respect to plane of seferences - - - - - - - - - - - - - - - - - - - - - - ---- - - ^ - - - - 3 - - - - - - Ñ ollut S Tacho Plane of reference .... . height = hay ho they hoy azero

Potential energy = mgh P.E a hi & hay ho they ho ho So, PE = (P.CA > (P.E)(P.E) > (PE), (P.E2| Kinetic energy - It is the energy of an object that it possesses due (KE) to its motion. | Kiti = a mual Ked v² at point A - body is at rest i.e. its v=0 30,(KE)A=0 As per Law of conservation of energy, total energy of an Isolated system remains constant.ie.. KE + PE = constant Now (PE)A > (PE)g (P.69€ (P.E.), (PF), So, (KE)A <UKE)& < (k€<(Kf)((KE), using above two conclusion we can say that - • At A - p.E - 4500 I (max.) KE = of (min.) • At B- PE = 3375 J KE = 1125 J Af C - PE = 0 J (min.) RiE = 4500 I (max.)

P.E= 11255 KE - 3375 J . At E. PE : 2250 J KE=2280J Атмишл- K'E-OJ PE=4500 J KE-ddSOJ P.E = 2250 K'E 11255 PE = 33755 KiE-337551 DE 21255 Les o KE: 4500 5 P.E = 0 J