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AI-তৈরি নমুনা উত্তর

Given data:
Mass of the car, $m = 2000\text{ kg}$
Angle of inclination, $\theta = 30^\circ$
Initial velocity, $u = 16\text{ ms}^{-1}$
Final velocity, $v = 0\text{ ms}^{-1}$
Stopping distance, $s = 40\text{ m}$
Acceleration due to gravity, $g = 9.8\text{ ms}^{-2}$

Using the kinematic equation for motion:
$v^2 = u^2 - 2as$
$0^2 = (16)^2 - 2 \times a \times 40$
$80a = 256$
$a = 3.2\text{ ms}^{-2}$

As the car moves down the incline, the component of gravity acting down the slope is $mg\sin\theta$, and the retarding force (kinetic friction along with braking resistance) $f_k$ acts up the incline to decelerate the car.
Applying Newton's second law along the incline:
$\Sigma F = f_k - mg\sin\theta = ma$
$f_k = m(a + g\sin\theta)$
$f_k = 2000\left(3.2 + 9.8\sin 30^\circ\right)$
$f_k = 2000\left(3.2 + 4.9\right) = 2000 \times 8.1 = 16200\text{ N}$

Therefore, the magnitude of the kinetic friction force acting on the car is $16200\text{ N}$ (or $16210\text{ N}$ taking $g = 9.81\text{ ms}^{-2}$).