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प्रश्न
The total energy of a body of mass 2 kg performing S.H.M. is 40 J. Find its speed while crossing the center of the path.
उत्तर १
Given:
Mass = m = 2 kg,
Energy = E = 40 J
The maximum speed of the body while crossing the path's centre (mean position) is Vmax, and the total energy is entirely kinetic energy.
∴ `1/2"mv"_"max"^2` = E
∴ vmax = `sqrt((2"E")/"m")=sqrt((2xx40)/2)` = 6.324 m/s
उत्तर २
Given: m = 2 kg, T.E. = 40 J
To find: Speed while crossing the mean position (vmax)
Formula: T.E. = `1/2"mv"_"max"^2`
Calculation:
From formula,
`"v"_"max" = sqrt((2 xx "T"."E".)/"m")`
= `sqrt((2 xx 40)/2)`
= `2sqrt10`
= 2 × 3.162
= 6.324 m/s
Speed of the particle while crossing the mean position is 6.324 m/s.
संबंधित प्रश्न
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