Physics, 19.07.2019 16:10 katelynalivia

A20 kg child is on a swing that hangs from 3.0-m-long chains. what is her maximum speed if she swings out to a 45° angle?

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Answer from: S4NCHEZ28

The maximum speed of the swing at an angle of 45° is 4.15 m/s.

Potential Energy, P. E = mgh Kinetic Energy, K. E = 0.5mv²

P.E = K.E

mgh = 0.5mv²

The vertical component of the swing = hCosθ

mghCosθ = 0.5mv²

20 × 9.8 × 3 × cos45° = 0.5 × 20 × v²

20 × 9.8 × 0.88 = 10v²

172.48 = 10v²

Divide both sides by 10 to isolate v²

v² = 172.48 / 10

v² = 17.248

v = √17.248

v = 4.15 m/s

Therefore, the maximum speed of the swing is 4.15 m/s

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Answer from: kornut5536

The maximum speed is 4.15 m/s.

Explanation:

To solve this problem we have to use the conservation of energy theorem, which says that the total energy in a system is constant.

In this case, we have kinetic energy (K) and potential energy (U), and the conservation refers to a continuous change between these two while the child is moving. When the child reaches his maximum amplitude on the swing, the kinetic energy is null and the potential energy is maximum. Then, when the child is in the middle of the movement, there the kinetic energy is maximum and the potential energy is null, because there's no height.

It's important to remember that kinetic energy depends on velocity and potential energy depends on position, height, we show this as equations:

$K=\frac{1}{2}mv^{2}$ ; where is the mass and is the velocity.

U=mgy ; where is the mass, is the gravity and is height.

Now, we analyse the movement. We are gonna use have of the whole movement of the swing, so, the initial movement is gonna be the middle, where the chain is perpendicular to the floor, and the final movement is gonna be where the chain forms 45°. The analysis is transformed into a equation.

So, according to the conservation of energy, the total energy at the beginning of the movement is the same at the final.

$E_{i}=E_{f} \\\frac{1}{2}mv_{i} ^{2}+mgy_{i} =frac{1}{2}mv_{f} ^{2}+mgy_{f}\\\frac{1}{2}mv_{i} ^{2}+0=0+mgy_{f}$

Now we solve for the initial velocity which is also the maximum speed, because we took the initial point where the chain is perpendicular to the ground, and as we said, in that point the potential energy is null and the kinetic energy is max.

$v_{i}=\sqrt{2gy_{f}}$

Now, we need to calculate $y_{f}$ :

$cos45=\frac{3-y_{f} }{3} \\3\frac{\sqrt{2}}{2}=3-y_{f} \\y_{f}=3- 3\frac{\sqrt{2}}{2}=0.88$

So, $y_{f}=0.88$ , now we replace this to find the initial or maximum velocity:

$v_{i}=\sqrt{2gy_{f}}$

$v_{i}=\sqrt{2(9.8)(0.88)}$

$v_{i}=4.15m/s$

Therefore, the maximum speed is 4.15 m/s.

Answer from: Quest

Sorry can’t u just tryna get points

Answer from: Quest

That would be the corect answer hope this

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A20 kg child is on a swing that hangs from 3.0-m-long chains. what is her maximum speed if she swing...