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