Answer:
[tex]N=21.690\,N[/tex]
Explanation:
The equation of equilibrium at the top of the vertical circle is:
[tex]\Sigma F = - N - m\cdot g = - m \cdot \frac{v^{2}}{R}[/tex]
The speed experimented by the car is:
[tex]\frac{N}{m}+g=\frac{v^{2}}{R}[/tex]
[tex]v = \sqrt{R\cdot (\frac{N}{m}+g) }[/tex]
[tex]v = \sqrt{(5\,m)\cdot (\frac{6\,N}{0.8\,kg} +9.807\,\frac{kg}{m^{2}} )}[/tex]
[tex]v\approx 9.302\,\frac{m}{s}[/tex]
The equation of equilibrium at the bottom of the vertical circle is:
[tex]\Sigma F = N - m\cdot g = m \cdot \frac{v^{2}}{R}[/tex]
The normal force on the car when it is at the bottom of the track is:
[tex]N=m\cdot (\frac{v^{2}}{R}+g )[/tex]
[tex]N = (0.8\,kg)\cdot \left(\frac{(9.302\,\frac{m}{s} )^{2}}{5\,m}+ 9.807\,\frac{m}{s^{2}} \right)[/tex]
[tex]N=21.690\,N[/tex]
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:
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:
[tex]N = F_c - mg[/tex]
But, [tex]F_c = \frac{mv^2}{r}[/tex]
[tex]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})}[/tex]
Substituting the given parameters into the formula, we have;
[tex]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 }[/tex]
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:
[tex]N = F_c + mg\\\\N = \frac{mv^2}{r} + mg\\\\N = m(\frac{v^2}{r} + g)[/tex]
Substituting the given parameters into the formula, we have;
[tex]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)[/tex]
Normal force, N = 21.68 Newton
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