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A trained sea lion slides from rest down a long
ramp into a pool of water. If the ramp is inclined at
an angle of 23° above the horizontal and the
coefficient of kinetic friction between the sea lion
and ramp is 0.26. Calculate the acceleration of the
sea lion as it slides down the ramp.​

Respuesta :

MathPhys

Answer:

1.5 m/s²

Explanation:

Draw a free body diagram.  There are three forces acting on the sea lion: gravity pulling down, normal force perpendicular to the ramp, and friction parallel to the ramp.

Sum of the forces perpendicular to the incline:

∑F = ma

N − mg cos θ = 0

N = mg cos θ

Sum of the forces parallel to the incline:

∑F = ma

mg sin θ − Nμ = ma

Substitute for N:

mg sin θ − (mg cos θ) μ = ma

g sin θ − g cos θ μ = a

a = g (sin θ − μ cos θ)

Given θ = 23° and μ = 0.26:

a = 9.8 (sin 23 − 0.26 cos 23)

a = 1.48

Rounded to two significant figures, the sea lion accelerates at 1.5 m/s².

Cricetus

The acceleration of the sea lion will be "1.484 m/s²".

Given:

  • Coefficient of Kinetic energy = [tex]0.26[/tex]
  • Angle = [tex]23^{\circ}[/tex]

By using the Newton's second equation, we get

→ [tex]mgSin \Theta-F_k = ma[/tex]

or,

→ [tex]mgSin \Theta-\mu mg Cos \Theta =ma[/tex]

→       [tex]g Sin \Theta-\mu g Cos \Theta =a[/tex]

hence,

The acceleration will be:

→ [tex]a = 9.8 Sin 23^{\circ} - 0.26\times 9.8\times Cos23^{\circ}[/tex]

     [tex]= 1.484 \ m/s^2[/tex]

Thus the above response is correct.

Learn more about newton's law here:

https://brainly.com/question/14050409

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