Respuesta :
Answer:
61.76 N.
Explanation:
Given the mass of the car, m = 1.60 kg.
The speed of the car, v = 12.0 m/s.
The radius of the circle, r = 5 m.
As car is moving in circular motion, so net force ( normal force + weight of the car) is equal to centripetal force enables the car to reamins in circular path.
Let N is the normal force.
So, [tex]N - mg = F_c[/tex]
[tex]N-mg=\frac{mv^2}{r}[/tex]
Now substitute the given values, we get
[tex]N-1.60kg\times9.8m/s^2=\frac{1.60kg\times(12.0m/s)^2}{5.0m}[/tex]
[tex]N=15.68+46.08[/tex]
N = 61.76 N.
Thus, the magnitude ofthe normal force exerted on the car by the walls is 61.76 N.
The magnitude of the normal force exerted on the car by the walls of the circle at point A is 16.96N
In order to get the magnitude of the normal force, we will use the formula:
[tex]N-mg=\frac{mv^2}{r}[/tex]
m is the mass
g is the acceleration due to gravity
v is the speed
r is the radius of the vertical circle
Given the following parameters
[tex]N-mg=\frac{mv^2}{r}\\N-1.60(9.8)=\frac{1.6\times 2^2}{5.0}\\N- 15.68=1.28\\N = 1.28+15.68\\N=16.96N[/tex]
Hence the magnitude of the normal force exerted on the car by the walls of the circle at point A is 16.96N
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