How much work should be done in the bicycle of mass 20kg to increase its speed from 2ms^-1 to 5ms^-1?( Ignore air resistance and friction)
Answers
Answered by
3
Work done is equal to change in kinetic energy of the bicycle
let v1= 2ms^-1
v2= 5ms^-1
Mass being constant, calculate the change in kinetic energy of the particle which will provide the work done
let v1= 2ms^-1
v2= 5ms^-1
Mass being constant, calculate the change in kinetic energy of the particle which will provide the work done
anish001:
thnz dear....im highly obliged
Answered by
3
Difference in the kinetic energy will he the work done by bicycle....
K. E = 1/2mv1^2 - 1/2mv2^2
K. E = 1/2 m *21
K. E= 10 * 21 = 210 j.....
I hope its correct.....
K. E = 1/2mv1^2 - 1/2mv2^2
K. E = 1/2 m *21
K. E= 10 * 21 = 210 j.....
I hope its correct.....
My pleasure.....
btw...its totally correct
Ohh actually I knew that I gotta be sure...
congz
But I was not confirm that's why I wrote I hope its is correct
that does'nt matter trying is the real achievement...rather than mitigating your own confidence
Hmm u r right
k. e = 1/2mv2^2 - 1/2mv1^2
best of luck for future and well tried☺
thnz mr.anonymous
Similar questions