Question

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​

Answer 1

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

     Similar link        

    https://brainly.in/question/5009654

Answer 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