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PhysicsW.E.P.Conservative force & potential energyMedium2 minPYQ_2020

PhysicsMediumnumerical

A disc of radius $0.1 m$ rolls without sliding on a horizontal surface with a velocity of $6 m s^{- 1}$ . It then ascends a smooth continuous track as shown in the figure. The height (in $m$ ) up to which it will ascend is : ( $g = 10 m s^{- 2}$ )

Answer:

2.70

Solution:

$\frac{1}{2} I w^{2} + \frac{1}{2} m v^{2} = m g h$

$\frac{1}{2} \frac{m r^{2}}{2} . \frac{v^{2}}{R^{2}} + \frac{1}{2} m v^{2} = m g h$

$\frac{v^{2}}{4} + \frac{v^{2}}{2} = g h$

$\frac{v^{2} + 2 v^{2}}{4} = g h$

$\frac{3}{4} \times 36 = 10 \times h$

$\frac{27}{10} = h \Rightarrow 2.7$

Stream:JEESubject:PhysicsTopic:W.E.P.Subtopic:Conservative force & potential energy

⏱ 2mℹ️ Source: PYQ_2020

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