Question

A student rides a bicycle in a circle at a constant speed and constant radius. A force diagram for the student-bicycle system is

shown in the figure above. The value for each force is shown in the figure. What is the acceleration of the student-bicycle system?
m
А
0
2
B
0.2"
C) 5
D
25"

Answer 1

The gravitational force that acts on the bicycle system is

`$F_(g) = 500 \ N$`

Now the force, that is the gravitational force is related to mass of the system and the acceleration due to gravity of the system, 'm' and 'g' respectively.

Therefore, we can write

`$F_g=mg$`

500 = m x 10 (since , g = 10 m/s-s)

∴ m = 50 kg

Now the net vertical force acting on the student bicycle system is 0. And the vertical acceleration of system is also 0. The total horizontal force acts to the right of the system. So by Newton's 2nd law of motion, we can write

`$F_f = ma$`

`$a=(F_f)/(m)$`

`$=(250)/(50)$`

Therefore
`$a= 5 \ m/s^2$`

Hence (C) is correct option.