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5.Work, Energy, Power and Collision
easy
The force constant of a weightless spring is $16 \,N/m$. A body of mass $1.0\, kg$ suspended from it is pulled down through $5\, cm$ and then released. The maximum kinetic energy of the system (spring + body) will be
A
$2 \times {10^{ - 2}}\,J$
B
$4 \times {10^{ - 2}}\,J$
C
$8 \times {10^{ - 2}}\,J$
D
$16 \times {10^{ - 2}}\,J$
Solution
(a)Max. K.E. of the system = Max. P.E. of the system
$\frac{1}{2}k{x^2}$= $ = \frac{1}{2} \times (16) \times {(5 \times {10^{ – 2}})^2} = 2 \times {10^{ – 2}}\,J$
Standard 11
Physics
