Physics, 10.03.2020 04:53 mika08

A small car with mass 0.800 kg travels at constant speed on the inside of the track that is a vertical circle with radius 5.00 m. 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 what is the speed of the car

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Answer from: zahriamarie10

a. The normal force on the car when it is at the bottom of the track is 21.68 Newton.

b. The speed of the car is 9.30 m/s.

Given the following data:

Mass of car = 0.800 kgRadius = 5.00 meterNormal force = 6.00 Newton

To find the normal force on the car when it is at the bottom of the track and the speed of the car:

At equilibrium, the normal force acting on the car at the top of the vertical circle is given by:

N = F_c - mg

But, $F_c = \frac{mv^2}{r}$

$N = \frac{mv^2}{r} - mg\\\\\frac{mv^2}{r} = N + mg\\\\v^2 = \frac{r(N + mg)}{m} \\\\v = \sqrt{r(\frac{N + mg}{m})}$

Substituting the given parameters into the formula, we have;

$v = \sqrt{\frac{6.00 + 0.800 \times 9.8}{0.800} \times 5}\\\\v = \sqrt{\frac{6.00 + 7.84}{0.800} \times 5}\\\\v = \sqrt{17.3 \times 5 } \\\\v = \sqrt{86.5 }$

Speed, v = 9.30 m/s

At equilibrium, the normal force acting on the car at the bottom of the vertical circle is given by:

$N = F_c + mg\\\\N = \frac{mv^2}{r} + mg\\\\N = m(\frac{v^2}{r} + g)$

Substituting the given parameters into the formula, we have;

$N = 0.800(\frac{9.30^2}{5} + 9.8)\\\\N = 0.800(\frac{86.49}{5} + 9.8)\\\\N = 0.800(17.30 + 9.8)\\\\N = 0.800(27.1)$

Normal force, N = 21.68 Newton

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Answer from: Liah34

$N=21.690\,N$

Explanation:

The equation of equilibrium at the top of the vertical circle is:

$\Sigma F = - N - m\cdot g = - m \cdot \frac{v^{2}}{R}$

The speed experimented by the car is:

$\frac{N}{m}+g=\frac{v^{2}}{R}$

$v = \sqrt{R\cdot (\frac{N}{m}+g) }$

$v = \sqrt{(5\,m)\cdot (\frac{6\,N}{0.8\,kg} +9.807\,\frac{kg}{m^{2}} )}$

$v\approx 9.302\,\frac{m}{s}$

The equation of equilibrium at the bottom of the vertical circle is:

$\Sigma F = N - m\cdot g = m \cdot \frac{v^{2}}{R}$

The normal force on the car when it is at the bottom of the track is:

$N=m\cdot (\frac{v^{2}}{R}+g )$

$N = (0.8\,kg)\cdot \left(\frac{(9.302\,\frac{m}{s} )^{2}}{5\,m}+ 9.807\,\frac{m}{s^{2}} \right)$

$N=21.690\,N$

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