A 100-g toy car is propelled by a compressed spring that starts it moving. the car follows a curved track. what is the final speed of the toy car if its initial speed is 2.00 m/s and it coasts up the frictionless slope, gaining 0.180 m in altitude.

Respuesta :

Answer:

0.687 m/s

Explanation:

Initial energy = final energy

1/2 mu² = mgh + 1/2 mv²

1/2 u² = gh + 1/2 v²

Given u = 2.00 m/s, g = 9.8 m/s², and h = 0.180 m:

1/2 (2.00 m/s)² = (9.8 m/s²) (0.180 m) + 1/2 v²

v = 0.687 m/s

The final speed of the toy car is about 0.687 m/s

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Further explanation

Let's recall Kinetic Energy Formula as follows:

[tex]\large {\boxed{Ek = \frac{1}{2}mv^2} }[/tex]

Ek = Kinetic Energy ( Joule )

m = mass of the object ( kg )

v = speed of the object ( m/s )

Let us now tackle the problem !

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Given:

mass of toy car = m = 100 g

initial speed of toy car = u = 2.00 m/s

initial height position of toy car = h₁ = 0 m

final height position of toy car = h₂ = 0.180 m

Asked:

final speed of toy car = v = ?

Solution:

We will use Conservation of Energy to solve this problem:

[tex]Ep_1 + Ek_1 = Ep_2 + Ek_2[/tex]

[tex]mgh_1 + \frac{1}{2} m u^2 = mgh_2 + \frac{1}{2} m v^2[/tex]

[tex]0 + \frac{1}{2}mu^2 = mgh_2 + \frac{1}{2} m v^2[/tex]

[tex]\frac{1}{2}u^2 = gh_2 + \frac{1}{2}v^2[/tex]

[tex]u^2 = 2gh_2 + v^2[/tex]

[tex]v^2 = u^2 - 2gh_2[/tex]

[tex]v = \sqrt {u^2 - 2gh_2}[/tex]

[tex]v = \sqrt {2.00^2 - 2(9.8)(0.180)}[/tex]

[tex]v \approx 0.687 \texttt{ m/s}[/tex]

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Learn more

  • Impacts of Gravity : https://brainly.com/question/5330244
  • Effect of Earth’s Gravity on Objects : https://brainly.com/question/8844454
  • The Acceleration Due To Gravity : https://brainly.com/question/4189441
  • Newton's Law of Motion: https://brainly.com/question/10431582
  • Example of Newton's Law: https://brainly.com/question/498822

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Answer details

Grade: High School

Subject: Physics

Chapter: Energy

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