Question

A small car with mass of 0.800 kg travels at a constant speed

Answer 1

Answer:

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