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Standard XII

Physics

Law of Conservation of Mechanical Energy

Fig. Shows th...

Question

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 $(g = 10 m s^{- 2})$

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

Decrease in height: h=14-7=7 m

Also KE at C =Loss in PE=mgh=140 J

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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=10ms−2)

The correct option is A 140 J Decrease in height: h=14-7=7 m Also KE at C =Loss in PE=mgh=140 J

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 $(g = 10 m s^{- 2})$

The correct option is A
140 J

Decrease in height: h=14-7=7 m

Also KE at C =Loss in PE=mgh=140 J