Recent content by sidrox

perfect! so the force is 40 * 0.4...and that by 5 gives you the power!

Work done by Friction + Work done by gravity = Final K.E. - Initial K.E. <i hope this is right!> now, u can work out the individual work done and find v.

The simplest way to look at it mathematically is \frac{dp}{dt}=F If the force along a direction is 0, the momentum change is 0, ie., the momentum is conserved. Note that it is only along THAT direction in which the momentum is conserved.

as long as \frac{mv^2}{r} < Friction or even equal to friction, the coin is going to stay. Whether or not it is in UCM the velocity is always tangential; so as soon as the centripetal force exceeds friction, the coin will fly off tangentially

i presumed it is friction that is providing the centripetal force...

and most are links to pdf files that can be downloaded :frown:!

i saw the list...but, they are really inactive...a tutorial on maybe differential calculus and integral calculus, covering all the types would be really useful...

Cant someone start a tutorial thread? Cant someone start like a tutorial thread? :confused: Looking at the potential here :bugeye: , if we could start a tutorial on a topic, maybe under a separate subdivision, and post there (not problems but different concepts that each one could bring out)...

m1 and m2 are two the two masses...

\frac{1}{2} m1 v^2 = Energy Released + \frac{1}{2} m2 v^2

lemme give u a tip...you need to remember that the mass is not constant and that the fuel used is continuously reducing the mass of the body...try to a get a start, use calculus and post how far you get...

Now the body comes to rest at t = \frac {2 \sqrt{v(0)}}{\alpha} Thus, i substituted the above expression for 't' in the expression for 'x'. and i have gotten the following value for x, but it is not one of the options. i got: x = \frac {8 \sqrt{v(0)}^3}{3 \alpha} Where...

okay...here is my working: 2 \sqrt{v} = - 2\sqrt{v(0)} - \frac {\alpha t}{2} On squaring, v = v(0) + \frac {\alpha^2 t^2}{4} + \sqrt{v(0)}\alpha t On writing v as \frac {dx}{dt} , and then integrating: x = 0 + \frac {\alpha^2 t^3}{12} + \frac { \sqrt{v(0)} \alpha t^2}{2} and i am...

is this true?? \frac{d}{dt} \-v(0) = 0 ? is that right?

i think that one confusion that needs to be cleared right away is that "your hand is in constant velocity because of the constant 1 N that is applied." F= ma The acceleration is constant, NOT the velocity...