Question

a small car with mass 0.670 kg travels at constant speed on the inside of a track that is a vertical circle with radius 5.00 m. part a if the normal force exerted by the track on the car when it is at the top of the track (point b) is 6.00 n, what is the normal force on the car when it is at the bottom of the track (point a)? express your answer with the appropriate units. f = value units

Explanation:

Step1: Analyze forces at top

At the top of the vertical - circular track, the net - force towards the center provides the centripetal force. The forces acting on the car are the normal force $N_{top}$ and the gravitational force $mg$. So, $F_{c}=N_{top}+mg$, where $m = 0.670\ kg$, $g = 9.8\ m/s^{2}$, and $N_{top}=6.00\ N$. First, find the centripetal force $F_{c}$:
$F_{c}=N_{top}+mg=6.00\ N+(0.670\ kg\times9.8\ m/s^{2})=6.00\ N + 6.566\ N=12.566\ N$.

Step2: Analyze forces at bottom

At the bottom of the vertical - circular track, the net - force towards the center is $F_{c}=N_{bottom}-mg$. Since the speed is constant, the centripetal force $F_{c}$ is the same as at the top. We know $F_{c}$ from Step 1, and we want to find $N_{bottom}$. Rearranging the formula for the net - force at the bottom gives $N_{bottom}=F_{c}+mg$.
Substitute $F_{c}=12.566\ N$ and $mg = 6.566\ N$ into the formula:
$N_{bottom}=12.566\ N+6.566\ N = 19.132\ N$.

Answer:

$19.1\ N$