CBSE Class 9 Answered
Q: |
A vehicle moving with a velocity of 2m/s can be stopped over a distance of 2 m.keeping the retarding force constant,if kinetic energy is doubles,what will be the distance covered before it comes to rest?
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Asked by Snehal Ambekar | 12 Mar, 2016, 11:28: AM
Expert Answer
Given:
u1 = 2 m/s
s = 2 m
Now v1 = 0
v12 – u12 = 2as
0 – 22 = 2×a×2
a = -1 ms-2
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 mu22 = 4m
u2 =
Let s’ = distance covered before it comes to rest
v22 – u22 = 2as'
02 – ()2
2×(-1)×s’
s' = 4 m
Answered by Faiza Lambe | 12 Mar, 2016, 06:01: PM
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