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To find the acceleration of the sled as it slides down a frictionless hill inclined at an angle of [tex]\( 28.0^\circ \)[/tex], we'll follow these steps:
1. Determine the angle in radians: To work with trigonometric functions in physics, angles should be in radians. For an angle [tex]\(\theta\)[/tex] in degrees, it can be converted to radians using the relation:
[tex]\[ \theta_{\text{rad}} = \theta_{\text{deg}} \times \frac{\pi}{180} \][/tex]
Given the incline angle [tex]\( \theta = 28.0^\circ \)[/tex]:
[tex]\[ \theta_{\text{rad}} \approx 28.0 \times \frac{\pi}{180} \approx 0.489 \, \text{radians} \][/tex]
(Note: The converted value ends up approximately as [tex]\( 0.4886921905584123 \, \text{radians} \)[/tex].)
2. Use the acceleration due to gravity: On any hill, the sled is accelerated by the component of gravitational force acting along the incline. The acceleration due to gravity ([tex]\( g \)[/tex]) is approximately [tex]\( 9.8 \, \text{m/s}^2 \)[/tex].
3. Calculate the component of gravitational force along the incline: The component of the gravitational force that leads to acceleration down the incline can be determined using the sine function:
[tex]\[ a = g \sin(\theta_{\text{rad}}) \][/tex]
Substituting the known values:
[tex]\[ a \approx 9.8 \times \sin(0.489) \][/tex]
4. Evaluate the sine function:
[tex]\[ \sin(0.489) \approx 0.469 \][/tex]
5. Compute the acceleration:
[tex]\[ a \approx 9.8 \times 0.469 \approx 4.60 \, \text{m/s}^2 \][/tex]
Hence, the acceleration of the sled as it slides down the frictionless hill inclined at [tex]\( 28.0^\circ \)[/tex] is approximately [tex]\( 4.60082131530173 \, \text{m/s}^2 \)[/tex].
1. Determine the angle in radians: To work with trigonometric functions in physics, angles should be in radians. For an angle [tex]\(\theta\)[/tex] in degrees, it can be converted to radians using the relation:
[tex]\[ \theta_{\text{rad}} = \theta_{\text{deg}} \times \frac{\pi}{180} \][/tex]
Given the incline angle [tex]\( \theta = 28.0^\circ \)[/tex]:
[tex]\[ \theta_{\text{rad}} \approx 28.0 \times \frac{\pi}{180} \approx 0.489 \, \text{radians} \][/tex]
(Note: The converted value ends up approximately as [tex]\( 0.4886921905584123 \, \text{radians} \)[/tex].)
2. Use the acceleration due to gravity: On any hill, the sled is accelerated by the component of gravitational force acting along the incline. The acceleration due to gravity ([tex]\( g \)[/tex]) is approximately [tex]\( 9.8 \, \text{m/s}^2 \)[/tex].
3. Calculate the component of gravitational force along the incline: The component of the gravitational force that leads to acceleration down the incline can be determined using the sine function:
[tex]\[ a = g \sin(\theta_{\text{rad}}) \][/tex]
Substituting the known values:
[tex]\[ a \approx 9.8 \times \sin(0.489) \][/tex]
4. Evaluate the sine function:
[tex]\[ \sin(0.489) \approx 0.469 \][/tex]
5. Compute the acceleration:
[tex]\[ a \approx 9.8 \times 0.469 \approx 4.60 \, \text{m/s}^2 \][/tex]
Hence, the acceleration of the sled as it slides down the frictionless hill inclined at [tex]\( 28.0^\circ \)[/tex] is approximately [tex]\( 4.60082131530173 \, \text{m/s}^2 \)[/tex].
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