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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?
Asked by Snehal Ambekar | 12 Mar, 2016, 11:28: AM
answered-by-expert Expert Answer

Given:

u1 = 2 m/s

s = 2 m

Now v1 = 0

v12 – u12 = 2as

begin mathsize 14px style rightwards double arrow end style 0 – 22 = 2×a×2

begin mathsize 14px style rightwards double arrow end style 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

begin mathsize 14px style rightwards double arrow end style u2 = begin mathsize 14px style 2 square root of 2 end style

Let s’ = distance covered before it comes to rest

v22 – u22 = 2as'

begin mathsize 14px style rightwards double arrow end style 02 – (begin mathsize 14px style 2 square root of 2 end style)2

begin mathsize 14px style rightwards double arrow end style 2×(-1)×s’

begin mathsize 14px style rightwards double arrow end style s' = 4 m

Answered by Faiza Lambe | 12 Mar, 2016, 06:01: PM
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