Answer:
(a) 0
(b) 10ML
(c) [tex]10ML(1 - cos(\theta))[/tex]
(d) [tex]10ML(1 + sin(\phi))[/tex]
Explanation:
(a) When hanging straight down. The child is at the lowest position. His potential energy with respect to this point would also be 0.
(b) Since the rope has length L m. When the rope is horizontal, he is at L (m) high with respect to the lowest swinging position. His potential energy with respect to this point should be
[tex]E_h = mgh = 10ML[/tex]
where g = 10m/s2 is the gravitational acceleration.
(c) At angle [tex]\theta[/tex] from the vertical. Vertically speaking, the child should be at a distance of [tex]Lcos(\theta)[/tex] to the swinging point, and a vertical distance of [tex]L - Lcos(\theta)[/tex] to the lowest position. His potential energy to this point would be:
[tex]E_{\theta} = mgh = 10M(L - Lcos(\theta)) = 10ML(1 - cos(\theta))[/tex]
(d) at angle [tex]\phi[/tex] from the horizontal. Suppose he is higher than the horizontal line. This would mean he's at a vertical distance of [tex]Lsin(\phi)[/tex] from the swinging point and higher than it. Therefore his vertical distance to the lowest point is [tex]L + Lsin(\phi) = L(1 + sin(\phi))[/tex]
His potential energy to his point would be:
[tex]E_{\phi} = mgh = 10ML(1 + sin(\phi))[/tex]
