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59. (II) The crate shown in Fig. 4-60 lies on a plane tilted at an angle A = 25.0° to the horizontal, with Mk 0.19. (a) Determine the acceleration of the crate as it slides down the plane. (b) If the crate starts from rest 8.15 m up along the plane from its base, what will be the crate's speed when it reaches the bottom of the incline?​

59 II The Crate Shown In Fig 460 Lies On A Plane Tilted At An Angle A 250 To The Horizontal With Mk 019 A Determine The Acceleration Of The Crate As It Slides D class=

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Explanation:

a) We need to write down first Newton's 2nd law as applied to the given system. The equations of motion for the x- and y-axes can be written as follows:

[tex]x:\;\;\;\;\;mg\sin 25° - \mu_kN = ma\;\;\;\;\;\;(1)[/tex]

[tex]y:\;\;\;\;\;N - mg\cos 25° = 0\;\;\;\;\;\;\;\;\;(2)[/tex]

From Eqn(2), we see that

[tex]N = mg\cos 25°\;\;\;\;\;\;\;(3)[/tex]

so using Eqn(3) on Eqn(1), we get

[tex]mg\sin 25° - \mu_kmg\cos 25° = ma[/tex]

Solving for the acceleration, we see that

[tex]a = g(\sin 25° - \mu_k\cos 25°)[/tex]

[tex]\;\;\;\;= 2.45\:\text{m/s}^2[/tex]

b) Now that we have the acceleration, we can now solve for the velocity of the crate at the bottom of the plane. Using the equation

[tex]v^2 = v_0^2 + 2ax[/tex]

Since the crate started from rest, [tex]v_0 = 0.[/tex] Thus our equation reduces to

[tex]v^2 = 2ax \Rightarrow v = \sqrt{2ax}[/tex]

[tex]v = \sqrt{2(2.45\:\text{m/s}^2)(8.15\:\text{m})}[/tex]

[tex]\;\;\;\;= 6.32\:\text{m/s}[/tex]

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