Given figure shows the vertical section of a frictionless surface. A block of mass 2 kg is released from the position A, its kinetic energy as it reaches the position C is

180 J

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140 J

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40 J

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280 J

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Solution

The correct option is D 140 J
As the there is no external force over it,
Energy of the body is conserved.
$\Rightarrow$ decrease in potential energy of the body = increase in kinetic energy of body.
intially KE=0.
kinetic energy at C=mgh=2(10)(14-7)=140J

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Fig. Shows the vertical section of a frictionless surface. A block of mass 2 kg is released from rest from position A; its KE as it reaches position C is $(g = 10 m s^{- 2})$

Q. 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)

Fig. Shows the vertical section of a frictionless surface. A block of mass 2 kg is released from rest from position A; its KE as it reaches position C is $(g = 10 m s^{- 2})$