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5.Work, Energy, Power and Collision
normal
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 $\mathrm{KE}=$ loss in $\mathrm{PE}$
$\mathrm{K}_{\mathrm{C}}-\mathrm{K}_{\mathrm{A}}=\mathrm{mgh}$
$\mathrm{K}_{\mathrm{A}}=0$
$\mathrm{K}_{\mathrm{C}}=2 \times 10 \times 7=140 \mathrm{\,J}$
Standard 11
Physics
