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Standard XII

Physics

Friction General Understanding

Figure shows ...

Question

Figure shows the vertical section of a frictionless surface. A block of mass 2kg is released from rest from position A. Its K.E as it reaches position C is (g = 10 m/s2)

140 J

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180 J

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120 J

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280 J

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Solution

The correct option is A 140 J
Applying WET from A to C,
$K_{A} + U_{A} = K_{C} + U_{C}$
$0 + m g (14) = K_{C} + m g (7)$
$\Rightarrow K_{C} = 7 m g$
$= 7 \times 2 \times 10$
$= 140 J$

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Figure shows the vertical section of a frictionless surface. A block of mass 2kg is released from rest from position A. Its K.E as it reaches position C is (g = 10 m/s2)

The correct option is A 140 JApplying WET from A to C, KA+UA=KC+UC 0+mg(14)=KC+mg(7) ⇒KC=7mg =7×2×10 =140J

The correct option is A 140 J
Applying WET from A to C,
$K_{A} + U_{A} = K_{C} + U_{C}$
$0 + m g (14) = K_{C} + m g (7)$
$\Rightarrow K_{C} = 7 m g$
$= 7 \times 2 \times 10$
$= 140 J$