An 8 kg block is pushed across a horizontal surface by a horizontally oriented force whose magnitude varies in time. The instantaneous power supplied by the pushing force to the block is P subscript p u s h end subscript open parentheses t close parentheses equals open parentheses 4 straight W over straight s cubed close parentheses t cubed If the block starts at rest at t = 0 s, what is the displacement (in m) of the block after 3 s?

Respuesta :

Answer:

4.5 m

Explanation:

We are given that

Mass of block,m=8 kg

[tex]P_push(t)=(4\frac{W}{s^3})t^3[/tex]

Apply work energy theorem

[tex]Wdt=d(K.E)[/tex]

Where [tex]W=4t^3[/tex]

Integrating on both sides

[tex]\int_{0}^{t} 4t^3dt=\int d(K.E)[/tex]

[tex]K.E=t^4[/tex]

[tex]v=\sqrt{\frac{2K.E}{m}}[/tex]

[tex]v=\sqrt{\frac{2t^4}{8}}=\frac{t^2}{2}[/tex]

[tex]\frac{ds}{dt}=\frac{t^2}{2}[/tex]

[tex]ds=\frac{1}{2}t^2dt[/tex]

integrating on both sides

[tex]s=\int_{0}^{t}\frac{1}{2}t^2 dt[/tex]

[tex]s=\frac{1}{6}t^3[/tex]

Substitute t=3 s

[tex]s=\frac{1}{6}(3)^3=4.5 m[/tex]

Hence, the displacement of the block after 3 s=4.5 m

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