A body of mass 2kg moving with a velocity of 2m/s. After sometime the velocity becomes 4m/s. Find the percentage change in kinetic energy.
Plzz answer my question as i am having exam tomorrow.
Answers
Answered by
6
change in kinetic energy=1/2×m(v²-u²)
1/2×2×(16-4)
1/2×2×12
12joule
percentage change in k.e.12/4×100=300%
1/2×2×(16-4)
1/2×2×12
12joule
percentage change in k.e.12/4×100=300%
Answered by
11
Initial kinetic energy
Ki = 1/2 m v^2 = 1/2×2×4= 4 J
Final kinetic energy
Kf = 1/2×2×16 = 16 J
% change in KE
=(Kf - Ki)/Ki ×100
= (16-4)/4 ×100
= 3×100= 300%
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