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Ber [7]
3 years ago
11

The 20 kg at angle of 53⁰ in an inclined plane is realsed from rest the coefficient of friction bn the block and the inclined pl

ane are ųs=0.3 and ųk=0.2
A) would the block move
B) if it moves what is its speed after it has descended a distance of 5m down the plane​
Physics
1 answer:
Scorpion4ik [409]3 years ago
7 0

<u>Given</u><u> </u><u>:</u><u>-</u>

  • A 20kg block at an angle 53⁰ in an inclined plane is released from rest .
  • \mu_s = 0.3 \ \& \ \mu_k = 0.2

<u>To </u><u>Find</u><u> </u><u>:</u><u>-</u>

  • Would the block move ?
  • If it moves what is its speed after it has descended a distance of 5m down the plane .

<u>Solution</u><u> </u><u>:</u><u>-</u>

For figure refer to attachment .

So the block will move if the angle of the inclined plane is greater than the <u>angle</u><u> of</u><u> </u><u>repose</u><u> </u>. We can find it as ,

\longrightarrow \theta_{repose}= tan^{-1}(\mu_s)

Substitute ,

\longrightarrow \theta_{repose}= tan^{-1}( 0.2)

Solve ,

\longrightarrow\underline{\underline{\theta_{repose}= 16.6^o }}

Hence ,

\longrightarrow\theta_{plane}>\theta_{repose}

<u>Hence</u><u> the</u><u> </u><u>block</u><u> will</u><u> slide</u><u> down</u><u> </u><u>.</u>

Now assuming that block is released from the reset , it's <u>initial</u><u> </u><u>velocity </u> will be 0m/s .

And the net force will be ,

\longrightarrow F_n = mgsin53^o - \mu_k N

Substitute, N = mgcos53⁰ ( see attachment)

\longrightarrow ma_n  = mgsin53^o - \mu_k mgcos53^o

Take m as common,

\longrightarrow\cancel{m }(a_n) = \cancel{m}( gsin53^o - \mu gcos53^o)

Simplify ,

\longrightarrow a_n = gsin53^o - \mu_k g cos53^o

Substitute the values of sin , cos and g ,

\longrightarrow a_n = 10( 0.79 - 0.2 (0.6))

Simplify ,

\longrightarrow a_n = 10 ( 0.79 - 0.12 ) \\\\ \longrightarrow a_n = 10 (0.67)\\\\ \longrightarrow \underline{\underline{a_n = 6.7 m/s^2}}

Now using the <u>Third </u><u>equation</u><u> </u><u>of</u><u> motion</u><u> </u>namely,

\longrightarrow2as = v^2-u^2

Substituting the respective values,

\longrightarrow2(6.7)(5) = v^2-(0)^2

Simplify and solve for v ,

\longrightarrow v^2 = 67 m/s\\\\\longrightarrow v =\sqrt{67} m/s \\\\\longrightarrow\underline{\underline{ v = 8.18 m/s }}

<u>Hence</u><u> the</u><u> </u><u>velocity</u><u> after</u><u> </u><u>covering</u><u> </u><u>5</u><u>m</u><u> </u><u>is </u><u>8</u><u>.</u><u>1</u><u>8</u><u> </u><u>m/</u><u>s </u><u>.</u>

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