a small mass attached to a string rotates on a frictionless table top if the tension in the string is increased by pulling the string causing the radius of the circular motion to decrease by a factor of 2,the kinetic energy of mass will?

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Solution

$a s w e k n o w m v r = c o n s \tan t m v r = m v' . \frac{r}{2} v' = 2 v t h e r e f o r e n e w k i n e t i c e n e r g y i s K . E' = \frac{1}{2} m (2 v)^{2} K . E = 4 \times K . E s o k i n e t i c e n e r g y w i l l b e c o m e 4 t i m e s .$

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a small mass attached to a string rotates on a frictionless table top if the tension in the string is increased by pulling the string causing the radius of the circular motion to decrease by a factor of 2,the kinetic energy of mass will?

as we know mvr = constant mvr=mv' .r2v' = 2vtherefore new kinetic energy is K.E'=12m(2v)2K.E= 4×K.Eso kinetic energy will become 4 times.

$a s w e k n o w m v r = c o n s \tan t m v r = m v' . \frac{r}{2} v' = 2 v t h e r e f o r e n e w k i n e t i c e n e r g y i s K . E' = \frac{1}{2} m (2 v)^{2} K . E = 4 \times K . E s o k i n e t i c e n e r g y w i l l b e c o m e 4 t i m e s .$