Answer:
The length of the driveway is 4.98 m
Explanation:
Given that m = mass of car = 2.1 × 10³ kg, d = length of driveway, f = frictional force = 4.0 × 10³ N, v = velocity of car at bottom of driveway = 3.8 m/s, u = speed of car at top of driveway = 0 m/s (since it starts at rest) and θ = slope of driveway = 20°
The component of the weight of the car along the driveway minus the frictional force = the net force. So,
mgsinθ - f = ma where a = acceleration of the car
a = gsinθ - f/m
substituting the variables, we have
a = 9.8 m/s²sin20° - 4.0 × 10³ N/2.1 × 10³ kg
a = 3.352 m/s² - 1.905 m/s²
a = 1.447 m/s²
a ≅ 1.45 m/s²
Using v² = u² + 2ad, we find the length of the driveway, d. So,
d = (v² - u²)/2a
Substituting the values of the variables, we have
d = (v² - u²)/2a
d = ((3.8 m/s)² - (0 m/s)²)/(2 × 1.45 m/s²)
d = 14.44 m²/s² ÷ 2.9 m/s²
d = 4.98 m
So, the length of the driveway is 4.98 m
The length of the driveway on which the car is moving is 5 m.
The given parameters;
The net horizontal force on the car is calculated as follows;
[tex]mgsin \theta - F = ma\\\\a = \frac{mgsin \theta - F}{m} \\\\a = \frac{(2100 \times 9.8 \times sin \ 20) - 4000}{2100}\\\\a = 1.45 \ m/s^2[/tex]
The length of the driveway is calculated as follows;
[tex]v^2 = u^2 + 2as\\\\s = \frac{v^2 - u^2 }{2a} \\\\s = \frac{3.8^2 - 0}{2\times 1.45} \\\\s = 5 \ m[/tex]
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