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A body of mass \(6 \mathrm{~kg}\) is acted upon by a force which causes a displacement in it given by \(x=\frac{t^2}{4}\) metre where \(t\) is the time in second. The work done by the force in 2 seconds is
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The correct answer is:
\(3 \mathrm{~J}\)
\(\begin{aligned}
& \text {Hints : } \mathrm{m}=6 \mathrm{~kg} \quad \mathrm{x}=\frac{\mathrm{t}^2}{4} \\
& \frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{v}=\frac{\mathrm{t}}{2} \quad \mathrm{v}(0)=0 ; \mathrm{v}(2)=\frac{2}{2}=1 \\
& \mathrm{~K}_{\mathrm{i}}=\frac{1}{2} \mathrm{~m}(0)^2=0 ; \mathrm{K}_{\mathrm{f}}=\frac{1}{2} \mathrm{~m}(1)^2=\frac{1}{2} \times 6 \times 1=3 ; \mathrm{W}=\mathrm{K}_{\mathrm{f}}-\mathrm{K}_{\mathrm{i}}=3-0=3 \mathrm{~J}
\end{aligned}\)
& \text {Hints : } \mathrm{m}=6 \mathrm{~kg} \quad \mathrm{x}=\frac{\mathrm{t}^2}{4} \\
& \frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{v}=\frac{\mathrm{t}}{2} \quad \mathrm{v}(0)=0 ; \mathrm{v}(2)=\frac{2}{2}=1 \\
& \mathrm{~K}_{\mathrm{i}}=\frac{1}{2} \mathrm{~m}(0)^2=0 ; \mathrm{K}_{\mathrm{f}}=\frac{1}{2} \mathrm{~m}(1)^2=\frac{1}{2} \times 6 \times 1=3 ; \mathrm{W}=\mathrm{K}_{\mathrm{f}}-\mathrm{K}_{\mathrm{i}}=3-0=3 \mathrm{~J}
\end{aligned}\)
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