The height of the loop is 12.3 m
Explanation:
We can solve this problem by applying the principle of conservation of energy. In fact, the total mechanical energy of the car (the sum of its potential energy + its kinetic energy) must be conserved in absence of friction. So we can write:
[tex]U_i +K_i = U_f + K_f[/tex]
where
:
[tex]U_i[/tex] is the initial potential energy, at the bottom
[tex]K_i[/tex] is the initial kinetic energy, at the bottom
[tex]U_f[/tex] is the final potential energy, at the top
[tex]K_f[/tex] is the final kinetic energy, at the top
We can rewrite the equation as
[tex]mgh_i + \frac{1}{2}mu^2 = mgh_f + \frac{1}{2}mv^2[/tex]
where:
m = 50 kg is the mass of the car
[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity
[tex]h_i = 0[/tex] is the initial height
, at ground level
u = 20 m/s is the initial speed of the car
[tex]h_f[/tex] is the maximum height reached by the car, so the heigth of the loop
v = 12 m/s is the final speed of the car at the top of the loop
By solving the equation for [tex]h_f[/tex], we find:
[tex]\frac{1}{2}mu^2 = mgh_f + \frac{1}{2}mv^2\\h_f = \frac{u^2-v^2}{2g}=\frac{20^2-12.6^2}{2(9.8)}=12.3 m[/tex]
Learn more about kinetic energy and potential energy here:
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