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
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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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