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2 years ago

Olivia is on a swing at the playground.

1 answer:

maks197457 [2]2 years ago

7 0

Olivia is on a swing at the playground. The kinetic energy is increasing and her potential energy decreasing at point X.

<h3>What is kinetic energy and potential energy?</h3>

The kinetic energy of an object is the ability to do work by virtue of its motion and potential energy is the ability to do work by virtue of its position.

At point W and Z, Olivia is at the maximum displacement from the mean position, where kinetic energy is zero and potential energy is maximum.

At point Y, it is approaching to increase its potential energy and decreasing kinetic energy. Opposite to this, at point X, kinetic energy is increasing and potential energy is decreasing.

Thus, the kinetic energy is increasing and her potential energy decreasing at point X.

Learn more about kinetic energy and potential energy.

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If the charge on the negative plate of the capacitor is 121 nano-Coulomb, how many excess electrons are on that plate? Write you

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n = 756.25 giga electrons

Explanation:

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3 years ago

A mover brings a box up the stairs in 10 seconds. If he applied a force of 20 N over a distance 10 m on the box, calculate the p

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3 years ago

Which transformation of energy takes place when a slingshot launches a stone? (PE stands for potential energy.) A. elastic PE to

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3 years ago

Read 2 more answers

The height of a typical playground slide is about 1.8 m and it rises at an angle of 30 ∘ above the horizontal.

Salsk061 [2.6K]

Answer:

$5.94\ \text{m/s}$

$1.7$

$0.577$

Explanation:

g = Acceleration due to gravity = $9.81\ \text{m/s}^2$

$\theta$ = Angle of slope = $30^{\circ}$

v = Velocity of child at the bottom of the slide

$\mu_k$ = Coefficient of kinetic friction

$\mu_s$ = Coefficient of static friction

h = Height of slope = 1.8 m

The energy balance of the system is given by

$mgh=\dfrac{1}{2}mv^2\\\Rightarrow v=\sqrt{2gh}\\\Rightarrow v=\sqrt{2\times 9.81\times 1.8}\\\Rightarrow v=5.94\ \text{m/s}$

The speed of the child at the bottom of the slide is $5.94\ \text{m/s}$

Length of the slide is given by

$l=h\sin\theta\\\Rightarrow l=1.8\sin30^{\circ}\\\Rightarrow l=0.9\ \text{m}$

$v=\dfrac{1}{2}\times5.94\\\Rightarrow v=2.97\ \text{m/s}$

The force energy balance of the system is given by

$mgh=\dfrac{1}{2}mv^2+\mu_kmg\cos\theta l\\\Rightarrow \mu_k=\dfrac{gh-\dfrac{1}{2}v^2}{gl\cos\theta}\\\Rightarrow \mu_k=\dfrac{9.81\times 1.8-\dfrac{1}{2}\times 2.97^2}{9.81\times 0.9\cos30^{\circ}}\\\Rightarrow \mu_k=1.73$

The coefficient of kinetic friction is $1.7$ .

For static friction

$\mu_s\geq\tan30^{\circ}\\\Rightarrow \mu_s\geq0.577$

So, the minimum possible value for the coefficient of static friction is $0.577$ .

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3 years ago