Search 75,786 tutors
0 0

## A roller-coaster car speeds down a hill past point A where R1 = 11.6 m and then rolls up a hill past point B where R2 = 15.2 m, as shown below.

A roller-coaster car speeds down a hill past point A where R1 = 11.6 m and then rolls up a hill past point B where R2 = 15.2 m, as shown below.

(a) The car has a speed of 20.0 m/s at point A. if the track exerts a normal force on the car of 2.08 104 N at this point, what is the mass of the car?

(b) What is the maximum speed the car can have at point B for the gravitational force to hold it on the track?

Dear Taliyah

You typed the normal force is 2.08 104 N, but I assuming this value is 2.08 x 10^4 [N].

(a) If you draw a free body diagram for car at point A, then N goes upward, mg goes downward and the remained force makes centripetal force to the center of circle.

Therefore, N - mg = Fc , and Fc = (m v^2 )/ r1

If you re-arrange the equation for m ,

m = N / ( g + v^2/r1 ) substitute given values to the equation gives

m = 469.71 kg ( very light car )

(b) you also need to draw a free body diagram. At point B, the normal force goes zero when the speed is maximum. If the speed is bigger than the maximum speed, then the car will fly away and there will be no normal force. In this situation, the weight is the same as the centripetal force.

mg = Fc = (m v^2 / r2 )

solve the equation gives v = √( r2 g ) = 12.21 m/s ; this is the maximum speed at B.

Hope this helps

If you have further question please let me know.