Question

Dr. John Schwartz takes all of the water resources students to Kings Dominion, where they eat donuts and ride the Yolcano Assume the 30 KN roller coaster starts from rest and is launched with electromagnetic motors along a horizontal section of track It takes 10 m to reach a speed of 32 m/s. UNITS NOTES Part IMAGES DISCUSS Description STATS HELP PREFERENCES Answer Chk History What force is applied to the roller coaster during the launch? (include units with answer) 12.5pts. 100% #tries: 0 Show Details A meter measures the energy input into the motors during the launch as 0.763 kW-hour. What is the efficiency of the motors? (%) #tries: 0 Show Details 12.5 pts. 100% After the launch, the roller coaster goes straight up, shooting out of the mountain. A minimum speed of 7.2 m/s is required at the top. Assuming no losses, how high above the initial elevation can the coaster go? (include units with answer) tries: Show Details 2 typy Hinta As the coaster comes down the hill to the same elevation at which it started, it loses 67400) of energy. What is the speed at the bottom of the hill? (include units with answer) 12.5pts.com tries: 0 Show Detail Hits29 12.5pt After the run, the roller coaster comes into the station with a speed of 0.58 times the speed at the end of the launch, what percent of energy was lost during the run? e s: Show Detail

Answer 1

a)

Mass of roller coaster

M=W/g =30000/9.81 =3058.1 kg

From Work energy theorem

Fd = (1/2)mV²

F*10 = (1/2)(3058.1)(32)²

F=156575 N

b)

Input Energy to motor

E_input=0.763*10³*3600 =2746800 J

Output Energy

E_output=(1/2)mV² =(1/2)(3058.1)(32)² = 1565749 J

efficiency

n=(E_out/E_in)*100 =(1565749/2746800)*100

n=57 %

c)

By Conservation of energy

(1/2)mV_i² = mgh +(1/2)mV_f²

(1/2)(32)² = 9.81*h + (1/2)(7.2)²

h=49.55 m

d)

From Work energy theorem

mgh +(1/2)mV_i² = E_Lost+(1/2)mV_f²

3058.1*9.81*49.55 +(1/2)(3058.1)(7.2)² = 67400 +(1/2)(3058.1)V_f²

V_f=31.303 m/s

e)

Now final speed

V_f=0.58*32 =18.56 m/s

Percentage of energy lost during the run is

% E_Lost = [(1/2)mV_i²-(1/2)mV_f²]/(1/2)mV_i²] = (32²-18.56²)/32²=0.6636

% E_Lost = 66.36 %