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Ugonna stands at the top of an incline and pushes a 100−kg crate to get it started sliding down the incline. The crate slows to a halt after traveling 1.50 m along the incline. (a) If the initial speed of the crate was 1.77 m/s and the angle of inclination is 30.0°, how much energy was dissipated by friction? (b) What is the coefficient of sliding friction?

Respuesta :

Answer:(a)891.64 N

(b)0.7

Explanation:

Mass of crate [tex]m=100 kg[/tex]

Crate slows down in [tex]s=1.5 m[/tex]

initial speed [tex]u=1.77 m/s[/tex]

inclination [tex]\theta =30^{\circ}[/tex]

From Work-Energy Principle

Work done by all the Forces is equal to change in Kinetic Energy

[tex]W_{friction}+W_{gravity}=\frac{1}{2}mv_i^2-\frac{1}{2}mv_f^2[/tex]

[tex]W_{gravity}=mg(0-h)=mgs\sin \theta [/tex]

[tex]W_{gravity}=-mgs\sin \theta[/tex]

[tex]W_{gravity}=-100\times 9.8\times 1.5\sin 30=-735 N[/tex]

change in kinetic energy[tex]=\frac{1}{2}\times 100\times 1.77^2=156.64 J[/tex]

[tex]W_{friction}=156.64+735=891.645[/tex]

(b)Coefficient of sliding friction

[tex]f_r\cdot s=W_{friciton}[/tex]

[tex]891.645=f_r\times 1.5[/tex]

[tex]f_r=594.43 N[/tex]

and [tex]f_r=\mu mg\cos \theta [/tex]

[tex]\mu 100\times 9.8\times \cos 30=594.43[/tex]

[tex]\mu =0.7[/tex]

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