pls help n.mastering physics.com/myct/itemView?assignmentproblemID-8131 7914&off
ID: 1791087 • Letter: P
Question
pls help
n.mastering physics.com/myct/itemView?assignmentproblemID-8131 7914&offsets; next CH 10 HW Problem 10.43 previous l 12 of 17 Inet » 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 10-m-high hill, then descends 20 m to the track's lowest point. You've What spring constant should you specity? Express your answer to two significant figures and include the appropriate units ing c k= 2.5x104 maximum of 2.0 m and that a loaded car will have a maximum mass of 450 kg For safety reasons the spring constant should be 15 % larger than the minimum needed for the car to just make it over the top. Correct 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. aliie Submit My Answers Give Up Incorrect; Try Again: 3 attempts remainingExplanation / Answer
A] The gravitational potential energy of the coaster on the top of the 10 m high hill (relative to its initial starting point) is:
PEg = m g h
PEg = (450 kg) (9.8 m/s^2) (10 m)
= 44100 J
To reach that height, the elastic potential energy stored in the spring must be the same, so:
PEg = PEe = k d^2 / 2
(44100 J) = k (2 m)^2 / 2
k = 44100*2/2^2
= 22050 N/m
Adding 14% to that value, you get:
K = 1.14 *(22050 N/m)
= 25137 N/m = 2.5 *10^4 N/m
B] When fully compressed, the elastic potential energy stored in the spring is:
PEe = K d^2 / 2
= (25137 N/m) (2 m)^2/ 2
= 50274 J
The difference in height between the starting point and the lowest point of the track is:
delta h = H - h
= (20 m) - (10 m)
= 10 m
So the initial gravitational potential energy of 330 kg coaster, relative to the lowest point, is
PEg = m g delta h
= (350 kg)* (9.8 m/s) *(10 m)
= 34300 J
The total energy of the coaster at its starting point (again, relative to the lowest point) is:
TE = PEe + PEg
= (50274 J) + (34300 J)
= 84574 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 = m V^2 / 2
(84574 J) = (350 kg) V^2 / 2
V = sqrt(84574*2/350)
= 21.98 m/s answer
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.