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Physics - W.E.P. Question with Solution | TestHub

PhysicsW.E.P.MiscellaneousMedium2 minQB

PhysicsMediumnumerical

A solid circular disc of mass $50 kg$ rolls along a horizontal floor so that its center of mass has a speed of $0.4 m s^{- 1}$ . The absolute value of work done on the disc to stop it is ______ $J$ .

Answer:

6.00

Solution:

Using work-energy theorem, we get

$W = ∆ K E$

$\Rightarrow W = 0 - (\frac{1}{2} m v^{2} + \frac{1}{2} I ω^{2})$

$\Rightarrow W = 0 - \frac{1}{2} m v^{2} (1 + \frac{K^{2}}{R^{2}})$

$= - \frac{1}{2} \times 50 \times 0 . 4^{2} (1 + \frac{1}{2}) = - 6 J$

Absolute work $= + 6 J$

Therefore, required value of $W = 6 J$

Stream:JEESubject:PhysicsTopic:W.E.P.Subtopic:Miscellaneous

⏱ 2mℹ️ Source: QB

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