To find the [tex]\( y \)[/tex]-component of the weight of the ball on an inclined plane, we need to follow these step-by-step procedures:
1. Understand the Problem:
- We have a ball with a mass of 5.20 kg.
- The ball is on a hill inclined at an angle of [tex]\( 15.0^\circ \)[/tex].
- We need to find the [tex]\( y \)[/tex]-component of the weight of the ball, which is the component perpendicular to the surface of the incline.
2. Identify Given Information:
- Mass ([tex]\( m \)[/tex]) = 5.20 kg
- Incline angle ([tex]\( \theta \)[/tex]) = [tex]\( 15.0^\circ \)[/tex]
- Acceleration due to gravity ([tex]\( g \)[/tex]) = 9.81 m/s[tex]\(^2\)[/tex]
3. Convert the Angle to Radians:
- Since trigonometric functions in most calculations assume angles in radians, we first convert [tex]\( 15.0^\circ \)[/tex] to radians.
[tex]\[
\theta_{\text{radians}} = 15.0^\circ \times \left(\frac{\pi}{180}\right) \approx 0.2617993877991494 \text{ radians}
\][/tex]
4. Calculate the y-component of the Weight:
- The weight ([tex]\( w \)[/tex]) of the ball is given by the formula [tex]\( w = m \times g \)[/tex].
[tex]\[
w = 5.20 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 50.972 \, \text{N}
\][/tex]
- The [tex]\( y \)[/tex]-component of the weight is the component acting perpendicular to the inclined plane. For a given angle [tex]\( \theta \)[/tex], this component can be found using the sine function:
[tex]\[
w_y = w \times \sin(\theta_{\text{radians}})
\][/tex]
Where:
[tex]\[
\sin(0.2617993877991494) \approx 0.258819045102521
\][/tex]
[tex]\[
w_y = 50.972 \times 0.258819045102521 \approx 13.20287712876979 \, \text{N}
\][/tex]
5. Conclusion:
- Therefore, the [tex]\( y \)[/tex]-component of the weight of the ball is approximately:
[tex]\[
w_y \approx 13.20287712876979 \, \text{N}.
\][/tex]
In summary, the [tex]\( y \)[/tex]-component of the weight of the 5.20 kg ball on a hill inclined at [tex]\( 15.0^\circ \)[/tex] is approximately [tex]\( 13.2 \, \text{N} \)[/tex].
