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5.Work, Energy, Power and Collision
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Figure shows the vertical section of frictionless surface. $A$ block of mass $2\, kg$ is released from the position $A$ ; its $KE$ as it reaches the position $C$ is ................ $\mathrm{J}$

A
$180$
B
$140$
C
$40$
D
$280$
Solution
Increase in $KE =$ loss in $PE$
$K_C -K_A = mgh$
$K_A = 0$
$K_C = 2 \times 10 \times 7 = 140\, J$
Standard 11
Physics
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