Question

what is the kinetic energy when 1/2 mass reduced and 1/2 speed reduced?

Answer 1

Answer:

${ \purple{the \: kinetic \: energy \: is \: reduced \: by \: 8 \: times \: from \: the \: initial.}}$

Explanation:

• ${ \huge{k}}o = \frac{1}{2} m {v}^{2}$
• now, m = m/2
• and, v=v/2
• now,
• ${ \huge{k}} = \frac{1}{2} \times \frac{m}{2} \times ({ \frac{v}{2}) }^{2}$
• $= \frac{m {v}^{2} }{16} = \frac{1}{8} \times \frac{1}{2} m {v}^{2} = {\huge{ \frac{1}{8} \times k}}o$
• ${ \orange{thank \: u......}}$
• ${ \red{please \: mark \: as \: brainliest}}$

Answer 2

Kinetic energy is One eight (1/8) when 1/2 mass reduced and 1/2 speed reduced

KE = (1/2) mv²

KE = Kinetic Energy

m = mass

v = velocity

Assume initial mass 2M and initial Velocity 2V

m = 2M

v = 2V

Hence Initial KE

KE = (1/2) 2M (2V)²

=> KE  = (1/2) 8MV²

1/2 mass reduced and 1/2 speed reduced

Hence Final mass = M  and Final velocity = V

Final KE =   (1/2)  M ( V)²

=>Final KE  = (1/2) MV²

=> Final KE  = (1/8)  (1/2) 8MV²

=> Final KE  = (1/8) Initial KE

Hence Kinetic energy is One eight (1/8) when 1/2 mass reduced and 1/2 speed reduced  

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