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Alexus [3.1K]
3 years ago
10

what is the kinetic energy of a 1 kilogram ball is thrown into the air with an initial velocity of 3 m/sec

Physics
1 answer:
s344n2d4d5 [400]3 years ago
3 0
Data:
m (mass) = 1 Kg
s (speed) = 3 m/s
Kinetic energy = ? (Joule)


Formula (Kinetic energy)
E_{k} =  \frac{m*s^2}{2}

Solving:
E_{k} = \frac{m*s^2}{2}
E_{k} =  \frac{1*3^2}{2}
E_{k} =  \frac{1*9}{2}
E_{k} =  \frac{9}{2}
\boxed{\boxed{E_{k} = 4.5\:J}}\end{array}}\qquad\quad\checkmark
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The height of a typical playground slide is about 1.8 m and it rises at an angle of 30 ∘ above the horizontal.
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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.

8 0
3 years ago
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