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A block of mass $10 \mathrm{~kg}$ is moving in $x$-direction with a constant speed of $10 \mathrm{~m} / \mathrm{s}$. It is subjected to a retarding force $F=0.1 x$ joule/metre during its travel from $x=20 \mathrm{~m}$ to $x=30 \mathrm{~m}$. Its final K.E. will be
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Verified Answer
The correct answer is:
$475 \mathrm{~J}$
Initial K.E., $E_1=\frac{1}{2} m v^2$
$$
=\frac{1}{2} \times 10 \times(10)^2=500 \mathrm{~J}
$$
At $x=20 \mathrm{~m}$, retarding force,
$$
F_1=0.1 \times 20=2 \mathrm{~N}
$$
At $x=30 \mathrm{~m}$, retarding force,
$$
F_2=0.1 \times 30=3 \mathrm{~N} \text {. }
$$
Average Retarding Force $F=\frac{F_1+F_2}{2}=\frac{2+3}{2}=2.5$
Work done by retarding force $=$ loss in K.E.
$$
=F \times s=2.5(30-20)=25 \mathrm{~J}
$$
Final K.E. $=E_1-\operatorname{loss}$ in K.E. $=500-25=475 \mathrm{~J}$
$$
=\frac{1}{2} \times 10 \times(10)^2=500 \mathrm{~J}
$$
At $x=20 \mathrm{~m}$, retarding force,
$$
F_1=0.1 \times 20=2 \mathrm{~N}
$$
At $x=30 \mathrm{~m}$, retarding force,
$$
F_2=0.1 \times 30=3 \mathrm{~N} \text {. }
$$
Average Retarding Force $F=\frac{F_1+F_2}{2}=\frac{2+3}{2}=2.5$
Work done by retarding force $=$ loss in K.E.
$$
=F \times s=2.5(30-20)=25 \mathrm{~J}
$$
Final K.E. $=E_1-\operatorname{loss}$ in K.E. $=500-25=475 \mathrm{~J}$
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