Recent content by Mimosapudica

I see! Thanks again for clearing this concept ^^

Ok so that’s why the force in the power equation is dependent on the man alone right ? Since it was derived from the work..

Ohhh right .. P= fv= W/t .. so W=fvt So again friction isn’t playing a part. Thank you! 😊

Ohh.. so power purely depends on the man ie the force he provides and the resulting velocity but not on any of the other forces that might act here. Am I right ? Just wondering tho, will work done by the man be dependent on the friction? Will we have to add the frictional force in that case?

Normal reaction gets canceled by the vertical component of T. (Tsin(theta)) Sorry, but if friction was provided, it would still be opposing the motion of the block, so won’t it reduce the net force acting in the direction of velocity?

Well yes, but only the component (horizontal)of the force in the direction of the velocity. The other component (vertical)cancels with the normal reaction.. does that mean there won’t be any friction if the normal reaction force gets canceled ?!

Isn’t frictional force equal to the normal reaction x coefficient of friction? Otherwise μk= tan (alpha) but this angle is the angle between the force and the normal reaction, which would be (90+ Theta)? I’m confused because if there is a coefficient of friction for the surface then friction...

Force along the horizontal would be T cos(theta) Frictional force (which is in the opposite direction )= μmg So net force in the direction of velocity = Tcos(theta)-μmg P= [Tcos(theta)-μmg]v But this is not so, the right answer is given to be Tvcos(theta). Why should we not consider the...

Oops.. latent heat is 80 cal/g .. So it doesn’t need conversation.. that was a careless error.. thank you.

I’m guessing the g to kg conversion has some problem, but shouldn’t we be converting it?

P=W/t W=Q= mL = (100x 80x 4.2)/1000 (kg x J/kg) = 33.6 t= 1 minute= 60s P= 33.6/60= 0.56 watt... but the answer provided is 560J/s...?