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loris [4]
2 years ago
13

A 16.0 kg child on roller skates, initially at rest, rolls 2.0 m down an incline at an angle of 20.0° with the horizontal. If th

ere is no friction between incline and skates, what is the kinetic energy of the child at the bottom of the incline?
Physics
1 answer:
enyata [817]2 years ago
7 0

The kinetic energy of the child at the bottom of the incline is 106.62 J.

The given parameters:

  • <em>Mass of the child, m = 16 kg</em>
  • <em>Length of the incline, L = 2 m</em>
  • <em>Angle of inclination, θ = 20⁰</em>

The vertical height of fall of the child from the top of the incline is calculated as;

sin(20) = \frac{h}{2} \\\\h = 2 \times sin(20)\\\\h = 0.68 \ m

The gravitational potential energy of the child at the top of the incline is calculated as;

P.E = mgh\\\\P.E = 16 \times 9.8 \times 0.68\\\\P.E = 106.62 \ J

Thus, based on the principle of conservation of mechanical energy, the kinetic energy of the child at the bottom of the incline is 106.62 J since no energy is lost to friction.

Learn more about conservation of mechanical energy here: brainly.com/question/332163

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Answer:

F=18.58\times 10^{-11}\ N

\theta=30.276^{\circ}

Explanation:

Given:

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<u>Now, gravitational force on particle 3 due to particle 1:</u>

F_{31}=G\frac{m_1.m_3}{r_{31}^2}

F_{31}=6.67\times 10^{-11} \times \frac{6.7\times 3.7}{4.2^2}

F_{31}=9.37\times 10^{-11}\ N

towards positive Y axis.

<u>gravitational force on particle 3 due to particle 2:</u>

F_{32}=G\frac{m_2.m_3}{r_{21}^2}

F_{32}=6.67\times 10^{-11} \times \frac{5.1\times 3.7}{2.8^2}

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<u>Now the net force</u>

F=\sqrt{F_{31}\ ^2+F_{32}\ ^2}

F=\sqrt{(10^{-11})^2(9.37^2+16.05^2)}

F=18.58\times 10^{-11}\ N

<em>For angle in counterclockwise direction from the +x-axis</em>

tan\theta=\frac{9.37\times 10^{-11}}{16.05\times 10^{-11}}

\theta=30.276^{\circ}

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