Suppose the roller coaster in Fig. 6-41 (h1 = 39 m, h2 = 12 m, h3 = 20) passes point 1 with a speed of?

1.70 m/s. If the average force of friction is equal to one fourth of its weight, with what speed will it reach point 2? The distance traveled is 50.0 m.

m/s

Suppose the roller coaster in Fig. 6-41 (h1 = 39 m, h2 = 12 m, h3 = 20) passes point 1 with a speed of?
Where is Fig. 6-41?
Reply:sum the energies equal to 0



starting KE

.5*m*1.7^2

loss of PE

m*g*(39-12)

loss due to friction

-m*g*0.25*50

KE at point 2

-.5*m*v^2



build the equation

starting KE

.5*m*1.7^2+m*g*(39-12)-

m*g*0.25*50-.5*m*v^2=0



solve for v

v=sqrt(1.7^2+2*9.81*(39-12)-

2*9.81*0.25*50)



17 m/s



____________________-