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Block $\mathrm{P}$ of mass $\mathrm{2 \mathrm{~kg}}$ slides down the surface and has a speed $\mathrm{20 \mathrm{~m} / \mathrm{s}}$ at the lowest point, $\mathrm{Q}$ where the local radius of curvature is $\mathrm{2 \mathrm{~m}}$ as shown in figure. Assuming $\mathrm{g=10 \mathrm{~m} / \mathrm{s}^2}$ , the normal force (inN) at $\mathrm{Q}$ is

Option: 1
$250 \mathrm{~N}$

Option: 2
$120 \mathrm{~N}$

Option: 3
$420 \mathrm{~N}$

Option: 4
$100 \mathrm{~N}$

Answers (1)

best_answer

$\mathrm{ \Sigma F_{\text {net }}=m a^r }$

$\mathrm{ N-m g=\frac{m v^2}{r} }$

$\mathrm{ N=m g+\frac{m v^2}{r} }$

$\mathrm{ =2 \times 10+2 \times \frac{(20)^2}{2}=20+400=420 \mathrm{~N} }$

Hence option 3 is correct.

Posted by

avinash.dongre

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