- #1
jiboom
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just a query about last part of a question.
a mass m hangs at end of light string and is raised vertically by an engine working at a constant rate kmg.at t=o,velocity =0 x=0 and x is measured in upwards direction.
now i have shown increase in total energy of the mass, when moved up a height h, due to this motion is
m(k^2lnB-ku)
it nows wants to deduce the time taken by considering the work done by the engine.
now i have said
work done is ∫kmg dx between 0 and h
so let x=t then dx=dt and x=o,t=o x=h t=t
then work done is kmgt
energy at bottom is 0 so energy gained by mass=work done by engine
so
m(k^2lnB-ku)=kmgt
ie gt=(klnB-u)
now I am sure this can't be right reasoning, but the final answer is correct.
firstly,is my expression for work done by engine correct?
secondly is my energy consideration correct? it says in my book in one place work done by force is change in ke and derives this then says later work done is equal to the change in mechanical energy,so which is the case?
the book even does a question of a mass going down a plane and calculates the velocity by working out work done by force displacing particle and equating this to KE,no mention of PE?
very confused
a mass m hangs at end of light string and is raised vertically by an engine working at a constant rate kmg.at t=o,velocity =0 x=0 and x is measured in upwards direction.
now i have shown increase in total energy of the mass, when moved up a height h, due to this motion is
m(k^2lnB-ku)
it nows wants to deduce the time taken by considering the work done by the engine.
now i have said
work done is ∫kmg dx between 0 and h
so let x=t then dx=dt and x=o,t=o x=h t=t
then work done is kmgt
energy at bottom is 0 so energy gained by mass=work done by engine
so
m(k^2lnB-ku)=kmgt
ie gt=(klnB-u)
now I am sure this can't be right reasoning, but the final answer is correct.
firstly,is my expression for work done by engine correct?
secondly is my energy consideration correct? it says in my book in one place work done by force is change in ke and derives this then says later work done is equal to the change in mechanical energy,so which is the case?
the book even does a question of a mass going down a plane and calculates the velocity by working out work done by force displacing particle and equating this to KE,no mention of PE?
very confused