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A $500 \mathrm{~kg}$ car takes a round turn of radius $50 \mathrm{~m}$ with a velocity of $36 \mathrm{~km} / \mathrm{h}$. The centripetal force is
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The correct answer is:
$1000 \mathrm{~N}$
The centripetal force required for a car (mass $m$ ) to take a sharp turn of radius $r$ with velocity $v$ is
$$
F=\frac{m v^{2}}{r}
$$
Given, $m=500 \mathrm{~kg}$,
$$
\begin{array}{l}
v=36 \mathrm{~km} / \mathrm{h}=36 \times \frac{5}{18}=10 \mathrm{~m} / \mathrm{s}, r=50 \mathrm{~m} \\
\text { Hence, } \quad F=\frac{500 \times(10)^{2}}{50}=1000 \mathrm{~N}
\end{array}
$$
$$
F=\frac{m v^{2}}{r}
$$
Given, $m=500 \mathrm{~kg}$,
$$
\begin{array}{l}
v=36 \mathrm{~km} / \mathrm{h}=36 \times \frac{5}{18}=10 \mathrm{~m} / \mathrm{s}, r=50 \mathrm{~m} \\
\text { Hence, } \quad F=\frac{500 \times(10)^{2}}{50}=1000 \mathrm{~N}
\end{array}
$$
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