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SashulF [63]
3 years ago
11

12. Prove that v' = u? + 2as.​

Physics
2 answers:
densk [106]3 years ago
7 0

Answer:

\boxed{ \boxed{ \orange{ \sf{see \: below}}}}

Explanation:

{  \underline{ \underline \bold{ \sf{ \blue{ {question \: }}}}}}

\sf{ \bold{ \underline{ \: prove \: that \:  {v}^{2}  =  {u}^{2}  + 2as}}}

\underline{ \underline{  { \bold{ \sf{ \purple{solution} }}}}}

Let us assume a body moving with an initial velocity ' u '. Let it's final velocity be 'v' after a time 't' and the distance travelled by the body be 's'. We already have ,

\sf{v = u + at} ⇒first equation of motion ( i )

\sf{s =  \frac{u + v}{2}  \times t} ⇒second equation of motion ( ii )

Putting the value of t from ( i ) in the equation ( ii )

{ v = u + at

or , at = v - u

or, t = v-u / a }

\sf{s =  \frac{u + v}{2}  \times  \frac{v - u}{a} }

\sf{or \: s =  \frac{ {v}^{2} -  {u}^{2}  }{2a} }

\sf{or \: 2as =  {v}^{2}  -  {u}^{2} }

\sf{ \boxed{ \bold{ \:   ∴  \:  \: \: {v}^{2}  =  {u}^{2}  + 2as}}} ⇒ forth equation of motion

Hope I helped!

Best regards!!

pashok25 [27]3 years ago
5 0
<h3><u>Correction</u><u>:</u><u>-</u></h3>

Prove that v² = u² + 2as .

<h3><u>Solution</u><u>:</u><u>-</u></h3>

From first equation of motion,

v = u + at

=> at = v - u

=> t = v - u / a

From the second equation of motion, we have

s = ut + 1/2at²

Putting the value of t in above equation , we get:

s = u ( v - u /a ) + 1/2 a ( v - u/a )²

s = uv - u² / a + a( v² + u² - 2uv / 2a²)

s = uv - u² / a + v² + u² - 2uv / 2a

s = 2uv - 2u² + v² + u² - 2uv / 2a

2as = v² - u²

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The question is incomplete. The complete question is :

In your job as a mechanical engineer you are designing a flywheel and clutch-plate system. Disk A is made of a lighter material than disk B, and the moment of inertia of disk A about the shaft is one-third that of disk B. The moment of inertia of the shaft is negligible. With the clutch disconnected, A is brought up to an angular speed ?0; B is initially at rest. The accelerating torque is then removed from A, and A is coupled to B. (Ignore bearing friction.) The design specifications allow for a maximum of 2300 J of thermal energy to be developed when the connection is made. What can be the maximum value of the original kinetic energy of disk A so as not to exceed the maximum allowed value of the thermal energy?

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Let M.I. of disk A = $I_0$

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After coupling, the angular velocity of both the disks will be equal to ω.

Angular momentum will be conserved.

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$I_0\omega_0 = I_0 \omega + 3I_0 \omega$

$I_0\omega_0 = 4I_0 \omega$

$\omega = \frac{\omega_0}{4}$

Now,

$K.E. = \frac{1}{2}I_0\omega^2+ \frac{1}{2}3I_0\omega^2$

$K.E. = \frac{1}{2}I_0\frac{\omega_0^2}{16}+ \frac{1}{2}3I_0\frac{\omega_0^2}{16}$

$K.E. = \frac{1}{2}I_0\omega_0^2 \left(\frac{1}{16}+\frac{3}{16}\right)$

$K.E. = \frac{1}{2}I_0\omega_0^2\times \frac{1}{4}$

$\Delta K = \frac{1}{2}I_0\omega_0^2 - \frac{1}{2}I_0\omega_0^2 \times \frac{1}{4} $

$2300=\frac{3}{4}\left(\frac{1}{2}I_0\omega_0^2\right)$

$\frac{1}{2}I_0\omega_0^2=2300 \times \frac{4}{3 } \ J $

Therefore, the maximum initial K.E. = 3066.67 J

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