441 J of work is required to slide a 15-kg ice block up a frictionless slope that is 20 m long.The incline of the slope with respect to horizontal is

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The work done to slide the ice block up the slope is equal to the gravitational potential energy gained, which is $mgh$, where $m$ is the mass, $g$ is the acceleration due to gravity (approximately $9.8 \, \text{m/s}^2$), and $h$ is the height gained. Since the slope is frictionless, all the work goes into increasing the gravitational potential energy.Given that the work done $W = 441 \, \text{J}$, and the mass $m = 15 \, \text{kg}$, we can express the height gained as $h = \frac{W}{mg}$.First, we find $h$:$$h = \frac{441}{15 \times 9.8} = \frac{441}{147} = 3 \, \text{m}$$Knowing the length of the slope ($l = 20 \, \text{m}$) and the height ($h = 3 \, \text{m}$), we can find the sine of the angle ($\theta$) of the slope using the right triangle formed by the slope, height, and the horizontal ground:$$\sin(\theta) = \frac{h}{l} = \frac{3}{20}$$To find the angle $\theta$, we take the inverse sine ($\sin^{-1}$) of $\frac{3}{20}$:$$\theta = \sin^{-1}\left(\frac{3}{20}\right) \approx 8.62^\circ$$Therefore, the incline of the slope with respect to the horizontal is approximately $8.62^\circ$.

441 J of Work Is Required to Slide a 15-Kg ∘\circ∘

441 J of Work Is Required to Slide a 15-Kg $\circ$