Question

1. -/2 points an acrobat swings on a 29.5 m long vine initially inclined at an angle of 40.0° with the vertical. (assume the vine has negligible mass.) (a) what is the acrobats speed (in m/s) at the bottom of the swing if she starts from rest? m/s (b) what is the acrobats speed (in m/s) at the bottom of the swing if instead she pushes off with a speed of 5.80 m/s? m/s resources read it

Explanation:

Step1: Find the height difference

The length of the vine is $L = 29.5$ m and the initial - angle with the vertical is $\theta=40.0^{\circ}$. The height difference $h$ between the initial and the lowest - point of the swing can be found using trigonometry. The vertical distance from the top of the vine to the initial position of the acrobat is $y = L\cos\theta$, and the vertical distance from the top of the vine to the bottom of the swing is $L$. So, $h = L - L\cos\theta$.
$h = L(1 - \cos\theta)=29.5\times(1 - \cos40.0^{\circ})\approx29.5\times(1 - 0.766)=29.5\times0.234 = 6.893$ m.

Step2: Use conservation of mechanical energy for part (a)

The initial mechanical energy $E_1$ of the acrobat is only potential energy since she starts from rest, $E_1=mgh$. At the bottom of the swing, the mechanical energy $E_2$ is only kinetic energy, $E_2=\frac{1}{2}mv^2$. By the conservation of mechanical energy $E_1 = E_2$ (neglecting air - resistance and the mass of the vine).
$mgh=\frac{1}{2}mv^2$. Canceling out the mass $m$ on both sides, we get $v=\sqrt{2gh}$.
Substituting $g = 9.8$ m/s² and $h = 6.893$ m, $v=\sqrt{2\times9.8\times6.893}=\sqrt{135.1028}\approx11.6$ m/s.

Step3: Use conservation of mechanical energy for part (b)

The initial mechanical energy $E_1$ is the sum of kinetic energy $K_1=\frac{1}{2}mu^2$ and potential energy $U_1 = mgh$, where $u = 5.80$ m/s. The final mechanical energy $E_2$ at the bottom of the swing is kinetic energy $K_2=\frac{1}{2}mv^2$.
By the conservation of mechanical energy $E_1 = E_2$, so $\frac{1}{2}mu^2+mgh=\frac{1}{2}mv^2$.
Canceling out the mass $m$ on both sides: $\frac{1}{2}u^2+gh=\frac{1}{2}v^2$.
We know that $u = 5.80$ m/s, $g = 9.8$ m/s², and $h = 6.893$ m.
First, calculate $\frac{1}{2}u^2=\frac{1}{2}\times(5.80)^2=\frac{1}{2}\times33.64 = 16.82$ and $gh=9.8\times6.893 = 67.5514$.
Then $\frac{1}{2}v^2=16.82 + 67.5514=84.3714$.
$v=\sqrt{2\times84.3714}=\sqrt{168.7428}\approx13.0$ m/s.

Answer:

(a) $11.6$ m/s
(b) $13.0$ m/s