Recent content by vandersmissen

It is the angle between the force and the displacement vector. the force is Mg since it is the mass times gravity the displacement would be found by calculating how much the block moves because of the force of block m falling. If block m falling travels a distance h, then block M moves a...

theta is the angle of the incline. I do not see a way to express this in other terms because the only variable of length I have is h .

well for block m , it would be cos 90 which would be 0 , but I was under the impression that you had to shift the axis so that there is actually a value calculated because if you have it as 0 no work is done. Which is true I suppose since it is moving up and down only. As for block M , you...

Well work is force times distance cos theta For the Work done by gravity it is mgh so would I set mgh = Mg[hsin(theta)]-f f = Mgcos(theta) so mh=M[hsin(theta)-cos(theta)] is that a step in the right direction ?

What I mean in the second part is that U1 + K1 , the initial gravitational and kinetic energy is equal to U2 + K2 the final graviational and kinetic energy. So since m starts at mgh and the system has an initial kinetic energy of (1/2)Mv^2 that is equal to the final which is...

Is anyone able to verify any of the above ?

Alright so for the first part k1= (1/2)mvi2 + (1/2)Mvi2 k2= (1/2)mvf2 + (1/2)Mvf2 \Deltak = k2-k1 so \Deltak = (1/2)(m+M)[vf2-vf2] That satisfies the first question. Then it asks for the work done by gravity. So we know that there is potential energy for the smaller block which is...

Homework Statement [PLAIN]http://rawrspace.com/physics.GIF Homework Equations v=v0+at x=x0+v0t+(1/2)at2 W=\DeltaK=K2-K1 The Attempt at a Solution For the first part of this problem I attempted it this way which I think is completely wrong. I rearranged the first two equations...

I had one additional question, It also asks Let M be the mass of the pulley in the previous question. If the effect of friction on the pulley is considered , how long does it take m2 to reach the floor? The mass is again released from rest and height h. To solve this would I follow the...

Thank you very much for the help.

Alright so We know that for mass 1 , T-m1g=m1a and for mass 2 , m2g-T=m2a The reason I set this up is because I was trying to make sure the force is positive and since m1 is moving upward, T is greater and on the otherside, since m2 is moving downward, it is greater than T , is that the correct...

Homework Statement It appears that the subscript is not working properly, please take m1 to means mass 1 and m2 to mean mass2 Atwood's machine consists of two masses connected by a string that passes over a pulley, as show in the figure. Consider the pulley to be massless and frictionless...

Homework Statement Find a vector-valued function f that traces out the given curve in the indicated direction. (a) Counterclockwise (b) Clockwise. 4x2+9y2=36 Homework Equations x2+y2=r2 cos2t+sin2t=1 The Attempt at a Solution From what I can determine, this is an ellipse. I...

I'm not too sure on that problem. What I think I can advise however is a few things. You are trying to maintain the speed, this means that final velocity and initial velocity are both 2 m/s which means that v-v_0=at 2m/s - 2m/s = at 0=at so there is no acceleration. If you take the...

Thank you for helping me understand where I went wrong.