Question

If the normal force exerted by the track on the car when it is at the top of the track (point BB) is 6.00 NN , what is the normal force on the car when it is at the bottom of the track (point AA)?

Answer 1

`N_A=21.68N`

Step-by-step explanation:

We already know that the wight in reference is gravitational force on an object:
`F_G=mg`.

Gravity=
`9.8m/s^2`

Weight of car=0.800kg

To solve for A, we apply Newton's second law of radial direction:

`\sum F_r_a_d=ma_R\\=mg+N_B\\`

Rewrite to calculate acceleration,
`a`:

`a_R=\frac {mg+N_B}{m}\\=(0.8kg+9.8m/s^2+6N)/(0.8kg)=17.3m/s^2`

Solving for Position A:

`\sum F_r_a_d=ma_R\\=N_A-mg\\N_A=ma_R+mg=0.8kg(17.3m/s^2+9.8m/s^2)\\=21.62N`