A block of mass m is released from top of a smooth fixed inclined plane of inclination - -Find out the speed of -160-the bIock at the -160-bottom- v-2gh- v-8gh- v-4gh- v-gh-

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Standard XII

Physics

Deducing Mathematically

A block of ma...

Question

A block of mass m is released from top of a smooth fixed inclined plane of inclination $θ$
Find out the speed of the bIock at the bottom.

v = \sqrt{g h}

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v = \sqrt{2 g h}

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v = \sqrt{4 g h}

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v = \sqrt{8 g h}

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Solution

The correct option is A $v = \sqrt{2 g h}$
Consider u and v be the initial and final velocity of the block.

Then, by applying equations of motion we can write:

$V^{2} = u^{2} + 2 a s$
$V^{2} = 0 + 2 (g s i n θ) \frac{h}{s i n θ} \Rightarrow V^{2} = 2 g h \Rightarrow V = \sqrt{2 g h}$

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A block of mass m is released from top of a smooth fixed inclined plane of inclination $θ$
Find out the speed of the bIock at the bottom.

The correct option is A $v = \sqrt{2 g h}$
Consider u and v be the initial and final velocity of the block.

Then, by applying equations of motion we can write:

$V^{2} = u^{2} + 2 a s$
$V^{2} = 0 + 2 (g s i n θ) \frac{h}{s i n θ} \Rightarrow V^{2} = 2 g h \Rightarrow V = \sqrt{2 g h}$

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A block of mass m is released from top of a smooth fixed inclined plane of inclination θFind out the speed of the bIock at the bottom.

The correct option is A v=√2ghConsider u and v be the initial and final velocity of the block.Then, by applying equations of motion we can write:V2=u2+2asV2=0+2(gsinθ)hsinθ⇒V2=2gh⇒V=√2gh

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A block of mass m is released from top of a smooth fixed inclined plane of inclination $θ$
Find out the speed of the bIock at the bottom.

The correct option is A $v = \sqrt{2 g h}$
Consider u and v be the initial and final velocity of the block.

Then, by applying equations of motion we can write:

$V^{2} = u^{2} + 2 a s$
$V^{2} = 0 + 2 (g s i n θ) \frac{h}{s i n θ} \Rightarrow V^{2} = 2 g h \Rightarrow V = \sqrt{2 g h}$