A block of mass $m$ is revolving in a smooth horizontal plane with a constant speed $v$ . If the radius of the circular path is $R$ , find the total constant force received by the block.

\frac{m v^{2}}{R}

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m \sqrt{\frac{v^{4}}{R^{2}} + 4 g^{2}}

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m \sqrt{\frac{v^{4}}{R^{2}} + g^{2}}

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Solution

The correct option is D $m \sqrt{\frac{v^{4}}{R^{2}} + g^{2}}$
$N = \sqrt{{(\frac{{m v}^{2}}{R})}^{2} + {(m g)}^{2}} N = m \sqrt{\frac{v^{4}}{R^{2}} + g^{2}}$

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A block of mass m is revolving in a smooth horizontal plane with a constant speed v. If the radius of the circular path is R, find the total constant force received by the block.

The correct option is D m√v4R2+g2N=√(mv2R)2+(mg)2N=m√v4R2+g2

A block of mass $m$ is revolving in a smooth horizontal plane with a constant speed $v$ . If the radius of the circular path is $R$ , find the total constant force received by the block.

The correct option is D $m \sqrt{\frac{v^{4}}{R^{2}} + g^{2}}$
$N = \sqrt{{(\frac{{m v}^{2}}{R})}^{2} + {(m g)}^{2}} N = m \sqrt{\frac{v^{4}}{R^{2}} + g^{2}}$