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A particle, starting from point A in the drawing (the height at A is 3.00 m), is projected down the curved runway. Upon leaving the runway at point B, the particle is traveling straight upward and reaches a height of 7.48 above the floor before falling back down. Ignoring friction and air resistance, find the speed of the particle at point A.

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cjrodriguez89

Answer:

vA = 9.37 m/s

Explanation:

  • Ignoring friction and air resistance, total mechanical energy must be conserved at any point, as follows:

        [tex]\Delta K + \Delta U = 0 (1)[/tex]

  • At the point A, the energy of the particle is in part gravitational potential energy, and in part kinetic energy.
  • So, we can write the following expressions for the initial energies:

       [tex]K_{iA} = \frac{1}{2} * m * v_{A} ^{2} (2)[/tex]

       [tex]U_{iA} = m*g*h = m*9.8 m/s2* 3.00 m (3)[/tex]

  • Since the total mechanical energy must be conserved at any point, we can choose as the final point, the one at which the ball, traveling straight upward, reaches to the maximum height before starting to fall down.
  • At this point, just barely before the ball changes direction, the speed of the ball is zero, so the final kinetic energy is zero too:

       [tex]K_{f} = 0 (4)[/tex]

  • The final gravitational potential energy, choosing as the zero reference level the ground, can be written as follows:

        [tex]U_{f} = m* g* H = m* 9.8 m/s2 * 7.48 m (5)[/tex]

  • Rearranging in (1) , we can write:

       [tex]K_{iA} + U_{iA} = K_{f} + U_{f} (6)[/tex]

  • Replacing by the values from (2), (3), (4) and (5) in (6), we get:

        [tex]\frac{1}{2} *m* v_{iA} ^{2} + m*9.8 m/s2*3.00 m = 0 + m*9.8 m/s2* 7.48 m (7)[/tex]

  • Simplifying and  rearranging, we finally can solve (7) for ViA, as follows:

        [tex]v_{iA} = \sqrt{2*9.8m/s2*4.48m}} = 9.37 m/s (8)[/tex]

Ignoring friction and air resistance, the speed of the particle at point A is; 9.37 m/s

The image of the particle projected down the curved runway has been attached.

  • This question is one that deals with conservation of energy. This means Initial energy is equal to final energy. Thus, the formula is;

¹/₂mv₂² + mgh₂ = ¹/₂mv₁² + mgh₁

Divide through by m to get;

¹/₂v₂² + gh₂ = ¹/₂v₁² + gh₁

Multiply through by 2 to get;

v₂² + 2gh₂ = v₁² + 2gh₁

where;

  • v₁ is speed at point A
  • v₂ is speed at point B
  • h₁ is height at point A
  • h₂ is height at point B
  • g is acceleration due to gravity

Now, at the top of the rise after point B, the velocity will be zero. Thus; v₂ = 0 m/s.

From the attached image;

h₁ = 3 m

h₂ = 7.48 m

Thus;

0² + (2 * 9.8 * 7.48) = v₁² + (2 * 9.8 * 3)

146.608 = v₁² + 58.8

v₁² = 146.608 - 58.8

v₁² =  87.808

v₁ = √87.808

v₁ = 9.37 m/s

Read more about conservation of energy at; https://brainly.com/question/11549071

Ver imagen AFOKE88

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