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Work Energy and Power question

2003 · Q154
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Work Energy and Power question

2003 · Q154

NEETPhysicsWork Energy and PowerMCQ+4 / −1
A particle moves along a circle of radius (20π)\left( {{{20} \over \pi }} \right)(π20​) m with constants tangential acceleration. If the velocity of vthe particle is 80 m/s at the end of the second revoluation after motion has begun, the tangential acceleration is
  1. A
    40 m/s2
  2. B
    640π\piπ m/s2
  3. C
    160π\piπ m/s2
  4. D
    40π\piπ m/s2
View written solutionFree

Correct answer: A

Wmg = Δ\Delta ΔK

⇒\Rightarrow⇒ -mgℓ\ell ℓ = ½ mv2 – ½ mu2
or, mv2 = m(u2 – 2gℓ\ell ℓ]
or, v=u2−2gℓj^v = \sqrt {{u^2} - 2g\ell } \widehat jv=u2−2gℓ​j​

u→=ui^\overrightarrow u = u\widehat iu=ui

∴v→−u→=u2−2gℓj^−ui^ \therefore \overrightarrow v - \overrightarrow u = \sqrt {{u^2} - 2g\ell } \widehat j - u\widehat i∴v−u=u2−2gℓ​j​−ui

∴∣v→−u→∣=[(u2−2gℓ)+u2]12 \therefore \left| {\overrightarrow v - \overrightarrow u } \right| = {\left[ {\left( {{u^2} - 2g\ell } \right) + {u^2}} \right]^{{1 \over 2}}}∴​v−u​=[(u2−2gℓ)+u2]21​

=2(u2−gℓ)= \sqrt {2\left( {{u^2} - g\ell } \right)}=2(u2−gℓ)​

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