an object of mass m and velocity v has kinetic energy = 200J .find the new kinetic energy
if the velocity of the object becomes double.
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Answered by
5
1/2 m v^2 = 200 (given)
Now the velocity of the particle becomes double
So new velocity = (2v )^2
New kinetic energy = 1/2 m (2v)^2
1/2 m 4v^2
let the new KE be x
original KE / New KE = 200/x
(1/2 m v^2 )/ (1/2 m 4v^2 ) = 200/ x
1/4= 200/x
x = 4 x 200 = 800
So
the new KE of the particle = 800 m/s
Now the velocity of the particle becomes double
So new velocity = (2v )^2
New kinetic energy = 1/2 m (2v)^2
1/2 m 4v^2
let the new KE be x
original KE / New KE = 200/x
(1/2 m v^2 )/ (1/2 m 4v^2 ) = 200/ x
1/4= 200/x
x = 4 x 200 = 800
So
the new KE of the particle = 800 m/s
maria9:
wlcm
u are studing in
Their is mistake in ur answer
what
1/4 = 200/x , x= 800 J
ya sorry
and also, velocity is not asked.... it's new K.E
thnks for correcting
velocity can't be determined !!!
Welcome :-)
Answered by
2
Given ;
K.E = 200 J
(1/2)mv^2 = 200
Now,
new velocity = v' = 2v
New kinetic energy =
K.E' = (1/2)mv'^2
= (1/2)×m×(2v)^2
= (1/2)mv^2 × 4
K.E' = 4 K.E
K.E' = 4×(200)
K.E' = 800 J
__________________
Hence the new kinetic energy will be increased by four times .
K.E = 200 J
(1/2)mv^2 = 200
Now,
new velocity = v' = 2v
New kinetic energy =
K.E' = (1/2)mv'^2
= (1/2)×m×(2v)^2
= (1/2)mv^2 × 4
K.E' = 4 K.E
K.E' = 4×(200)
K.E' = 800 J
__________________
Hence the new kinetic energy will be increased by four times .
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