Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

Problem 10.43 Part A You have been hired to design a spring-launched roller coas

ID: 1783256 • Letter: P

Question

Problem 10.43 Part A You have been hired to design a spring-launched roller coaster that will carry two passengers per car. The car goes up a 12-m-high hill, then descends 17 m to the track's lowest point. You ve determined that the spring can be compressed a maximum of 2.4 m and that a loaded car will have a maximum mass of 450 kg. For safety reasons, the spring constant should be 10 % larger than the minimum needed for the car to just make it over the top. What spring constant should you specify? Express your answer to two significant figures and include the appropriate units. | Value Units Submit My Answers Give Up Part B What is the maximum speed of a 350 kg car if the spring is compressed the full amount? Express your answer to two significant figures and include the appropriate units vValue Units Submit My Answers Give Up

Explanation / Answer

A] Energy from the spring = Energy absorbed by roller coaster to be able to make it to the top of a 12-meter hill

0.5kx^2 = mgh

0.5*k*2.4^2 = 450*9.8*12

k = 450*9.8*12/(0.5*2.4^2)

= 18375 N/m

Since it is required that the spring constant must be 10% larger than the minimum required, then

k = 18375 + 10*18375/100

= 1.1*18375

= 20212.5 N/m

b] By Conservation of energy,

TE = PEe + PEg

where, TE is total energy of the coaster at its starting point

PEe is  elastic potential energy stored in the spring

PEg is initial gravitational potential energy of 350 kg coaster, relative to the lowest point

TE = 0.5kx^2 + mg delta h

= 0.5*20212.5*2.4^2 + 350*9.8*( 17 - 12)

= 75362 J

At the lowest point of the track, all that energy is converted to kinetic energy, so the speed at that point will be

TE = KE = 75362 J

0.5mV^2 = 75362

V = sqrt(75362*2/350)

= 20.75 m/s answer