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ruert.meybiusi en/student/mainfr.uni G Google Chrome Web Stor... Paycheck PANET I Watch Descendan. PANET 1 s OHistory (1) YouTube Other Bookmarks LUS WileyPLUS MywitlevS Hele 1 Centast Us1 e Halliday, Fundamentals of Physics, 10e Grossmont College Physics (PHYC-140-240-241) & Practice Assignment Gradebook ORION Downloadable eTextbook Assignment MESSAGE MY INSTRUCTOR MEXT FULL SCREEN PRENTER VERSION SACK CES Chapter 08, Problem 009 19 In the figure, a frictionless roller coaster car of mass m 828 kg tops the first hill with speed va 15.0 m/s at height h 39.0 m. What is the speed of the car at (a) point A, (b) point 8, and (e) point C? (d) How high will the car go on the last hill, which is too high for it to cross? Use g-9.81 m/s First hill- Units (a) Number Units (b) Number (c) Number units Units (d) Number By accessing this Question Assistance, you will learn while you earn points based on the Point Potential Policy set by your instructor. SUBHIY ANSWwER SAVE FOR LATER Question Attempts: 0 of 10 used Earn Maximum Points available only if vou Version 4 24 17 ns. Inc. All Rights Reserved. A Division of John Wilev A Sons. Inc 0 2000-2019 John Wley Privacy Policy 1

Answer 1

here,

mass of car , m = 828 kg

v0 = 15 m/s

initial height , h = 39 m

a)

the height of point A ,hA = 39 m

using conservation of mechanical energy

it has same kinetic energy

so , the speed at point A is 15 m/s

b)

the height of point B , hB = h/2 = 19.5 m

using conservation of mechanical energy

KEa + PEa = KEb + PEb

0.5 * m * vA^2 + m * g * hA = 0.5 * m * vB^2 + m * g * hB

0.5 * 15^2 + 9.81 * 39 = 0.5 * vB^2 + 9.81 * (19.5)

solving for vB

vB = 24.6 m/s

the speed at point B is 24.6 m/s

c)

the height of point C , hC = 0 m

using conservation of mechanical energy

KEa + PEa = KEc + PEc

0.5 * m * vA^2 + m * g * hA = 0.5 * m * vC^2 + m * g * hc

0.5 * 15^2 + 9.81 * 39 = 0.5 * vc^2 + 0

solving for vc

vc = 31.5 m/s

the speed at point c is 31.5 m/s

d)

the maximum height , hm = vc^2 /2g

hm = 31.5^2 /(2*9.81) m = 50.5 m