Answer:
d = 27.7 m
Explanation:
Here the car is driving on the inclined plane
So here we can say that work done by the gravity and work done by friction is equal to change in kinetic energy of the system
So here we can write it as
[tex]mgsin\theta \times d - F_f \times d = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2[/tex]
now we have
m = 1700 kg
[tex]v_f = 0[/tex]
[tex]v_i = 21 m/s[/tex]
[tex]F_f = 1.5 \times 10^4 N[/tex]
[tex](1700)(9.81)sin5 (d) - (1.5\times 10^4)d = 0 - \frac{1}{2}(1700)(21^2)[/tex]
[tex]1453.5 d - (1.5\times 10^4)d = -374850[/tex]
[tex]d = 27.7 m[/tex]
The distance of the car's skid before stopping is 27.2 m.
The given parameters;
The distance of the car's skid before stopping is determined by applying the principle of conservation of energy.
work-done by friction = change in kinetic energy
[tex]mgsin(\theta) d - F_kd = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2\\\\(1700\times 9.8\times sin(5))d \ - \ 15000d = \frac{1}{2}\times 1700 \times 0^2\ \ -\ \ \frac{1}{2}\times 1700\times 21^2\\\\1449.42d \ - \ 15000d = - 374,850\\\\-13,550.58d = -374,850\\\\d = \frac{374,850}{13,550.58} \\\\d = 27.7 \ m[/tex]
Thus, the distance of the car's skid before stopping is 27.2 m.
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