Physics, asked by mehtabsinghwarn, 10 months ago

the distance covered by a body to come to rest when it is moving with a speed of 4 metre per second is S when a retarding force f is applied if the kinetic energy is doubled the distance covered by it to come to rest for the same retarding force f is​

Answers

Answered by knjroopa
11

Explanation:

Given The distance covered by a body to come to rest when it is moving with a speed of 4 metre per second is S when a retarding force f is applied if the kinetic energy is doubled the distance covered by it to come to rest for the same retarding force f is

  •                                         s = - 16 m / F
  • Mass of the body is m
  • We know that F = ma  
  • Or a = F/m
  • Equation of motion is
  • v^2 – u^2 = 2as
  • 0 – 4^2 = 2 x F/m x s
  • -16 = 2 F/m x s
  •  s = - 8m / F
  • K E = 1/2 m v^2
  •       = 1/2 x m x 4^2
  •       = 8 m is initial K E
  • Now 2 K E = 16 m
  •  So 2 x 1/2 m v^2 = 16 m
  •  So v^2 = 32
  • Or v = √32
  • Now v^2 – u^2 = 2 as
  •       0 – 32 = 2 x F/m x s
  • So s = - 16 m / F

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Answered by pavit15
2

Answer:

Explanation:

Given The distance covered by a body to come to rest when it is moving with a speed of 4 metre per second is S when a retarding force f is applied if the kinetic energy is doubled the distance covered by it to come to rest for the same retarding force f is

                                       s = - 16 m / F

Mass of the body is m

We know that F = ma  

Or a = F/m

Equation of motion is

v^2 – u^2 = 2as

0 – 4^2 = 2 x F/m x s

-16 = 2 F/m x s

s = - 8m / F

K E = 1/2 m v^2

     = 1/2 x m x 4^2

     = 8 m is initial K E

Now 2 K E = 16 m

So 2 x 1/2 m v^2 = 16 m

So v^2 = 32

Or v = √32

Now v^2 – u^2 = 2 as

     0 – 32 = 2 x F/m x s

So s = - 16 m / F

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