The vehicle moving with a velocity of 2 m/s can be stopped over a distance of 2m Keeping the retarding force constant, if the kinetic energy is doubled, what is the distance travelled before it comes to rest ?
Answers
Answered by
14
u1 = 2 m/s
s = 2 m
Now v1 = 0
v12 – u12 = 2as
=> 0 – 22 = 2×a×2
=> a = -1 ms-2
The Kinetic energy of the body,
K1= ½ m(2)2
= 2m
Now if Kinetic energy is doubled then,
K2 = 4m
Let velocity of the body become u2
½ mu2 2 = 4m
=> u2 = 2√2
Let s’ = distance covered before it comes to rest
v22 – u22 = 2as'
=> 02 – (2√2)2
= 2×(-1)×s’
=> s' = 4 m
s = 2 m
Now v1 = 0
v12 – u12 = 2as
=> 0 – 22 = 2×a×2
=> a = -1 ms-2
The Kinetic energy of the body,
K1= ½ m(2)2
= 2m
Now if Kinetic energy is doubled then,
K2 = 4m
Let velocity of the body become u2
½ mu2 2 = 4m
=> u2 = 2√2
Let s’ = distance covered before it comes to rest
v22 – u22 = 2as'
=> 02 – (2√2)2
= 2×(-1)×s’
=> s' = 4 m
Answered by
6
Answer:
new distance = s' = 4m
Explanation:
here's your anwser..(✯ᴗ✯)(✿^‿^)(✯ᴗ✯)(✿^‿^)
Attachments:
Similar questions