Answer:
a) 7.27 m/s
b) 0.676 m
Explanation:
From the law of conservation of mechanical energy, the potential energy converts to kinetic energy at the bottom. So, the lowest point of the swing, the girl would have maximum speed.
[tex]mg(h'-h) = 0.5 m v^2\\v=\sqrt{2g(h'-h)}[/tex]
Substitute the values:
[tex]v= \sqrt{2\times 9.8\times (3.4-0.7)}\\v = 7.27 m/s[/tex]
If the girl has speed half the maximum speed i.e. 3.64 m/s, then she will be:
[tex]h =\frac{v^2}{2g}\\h= \frac{3.64^2}{2\times 9.8}\\h=0.676 m[/tex]
Girl will be 0.676 m above the ground.
This question involves the concepts of the law of conservation of energy, kinetic energy, and potential energy.
a) The maximum speed of the girl is "7.28 m/s".
b) The height at which the girl will be moving at a speed that is half of her maximum speed is "1.33 m".
a)
According to the law of conservation of energy:
Gain in Potential Energy = Loss in Kinetic Energy
[tex]mg\Delta h=\frac{1}{2}mv_{max}^2\\\\v_{max}=\sqrt{2g\Delta h}[/tex]
where,
v_max = maximum speed = ?
g = acceleration due to gravity = 9.81 m/s²
Δh = change in height of the girl = 3.4 m - 0.7 m = 2.7 m
Therefore,
[tex]v_{max}=\sqrt{2(9.81\ m/s^2)(2.7\ m)}\\v_{max}=7.28\ m/s[/tex]
Hence, the total energy will be:
E = [tex]\frac{1}{2}mv_{max}^2 = \frac{1}{2}m(7.28\ m/s)^2[/tex]
E = 26.5 m
b)
Now, applying the law of conservation of energy to the point where the speed of the girls is half of the maximum speed:
[tex]E = P.E + K.E\\26.5\ m = mg\Delta h + \frac{1}{2}mv^2\\\\26.5 = (9.81)(h - 0.7\ m)+\frac{1}{2}(\frac{7.28\ m/s}{2})^2\\\\9.81\ h+6.87=26.5-6.62\\\\h=\frac{26.5-6.62-6.87}{9.81}[/tex]
h = 1.33 m
Learn more about the law of conservation of energy here:
brainly.com/question/20971995?referrer=searchResults
The attached picture explains the law of conservation of energy.
