ICSE Class 10 Answered
Speed of particle increases by 20%.How many percent its kinetic energy increases
Asked by lovemaan5500 | 26 Dec, 2017, 11:23: AM
let the initial velocity be V
20 % increase will be V + 20% of V
= 1.2V
KEi = 1/2 m v2
KEf = 1/2 m (1.2v)2
= 1.44 x 1/2 m v2
increase in KE= KEf - KEi = 1.44 x 1/2 m v2 - 1/2x m v2 = 0.44 x 1/2 m v2
percent increase in kinetic energy
![begin mathsize 20px style equals fraction numerator i n c r e a s e space i n space K E over denominator K E subscript i end fraction space equals space fraction numerator 0.44 space x 1 divided by 2 space m space v squared over denominator 1 divided by 2 space m space v squared end fraction space equals space 0.44 space end style](https://images.topperlearning.com/topper/tinymce/cache/ee62272e346f46dc57c80c555f7dfbf3.png)
Answer is 44 %
Answered by Gajendra | 27 Dec, 2017, 16:49: PM
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