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a 115 kg car is on a hill that is inclined at 37.0^{\\circ}. what is th…

Question

a 115 kg car is on a hill that is inclined at 37.0^{\circ}.

what is the x-component of the weight of the car?

w_{x} = ? n

Explanation:

🆕 New Concept Discovered: Forces on an Inclined Plane
Breaking gravity into components parallel and perpendicular to a slope.

Step 1: Calculate the total weight of the car

The total weight \( w \) of an object is the force of gravity acting on its mass. We calculate this by multiplying the mass \( m \) by the acceleration due to gravity \( g \approx 9.80\text{ m/s}^2 \):

$$ w = m \cdot g $$
$$ w = 115\text{ kg} \times 9.80\text{ m/s}^2 = 1127\text{ N} $$

Step 2: Find the x-component of the weight

When an object is on an inclined plane, we typically set up our coordinate system so that the x-axis runs parallel to the slope and the y-axis is perpendicular to it.

The component of gravity pulling the car down the slope (the x-component, \( w_x \)) is determined using the sine of the incline angle \(\theta\):

$$ w_x = w \cdot \sin(\theta) $$

Substitute the values into the formula:

$$ w_x = 1127\text{ N} \times \sin(37.0^\circ) $$
$$ w_x \approx 1127\text{ N} \times 0.6018 $$
$$ w_x \approx 678.2\text{ N} $$

Rounding to three significant figures (matching the given values \(115\text{ kg}\) and \(37.0^\circ\)):

$$ w_x \approx 678\text{ N} $$

Answer:

678