tag:blogger.com,1999:blog-5997147895880585179.post3325936394280639440..comments2024-03-17T18:54:13.159+05:30Comments on Solutions to I E Irodov - Physical Fundamentals of Mechanics: Irodov Problem 1.68Unknownnoreply@blogger.comBlogger6125tag:blogger.com,1999:blog-5997147895880585179.post-73427468853486771262012-07-10T10:15:29.613+05:302012-07-10T10:15:29.613+05:30i must say you have done most brilliant job on net...i must say you have done most brilliant job on net. I could have never guessed that this is a case of variable acceleration on my own. Thanks.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-5997147895880585179.post-52203204965954519392011-06-11T18:23:28.495+05:302011-06-11T18:23:28.495+05:30Yes indeed there were friction what you are saying...Yes indeed there were friction what you are saying is correct - but the problem says "smooth surface" implying that there is no frictionKrishna Kant Chintalapudihttps://www.blogger.com/profile/07024636005277277405noreply@blogger.comtag:blogger.com,1999:blog-5997147895880585179.post-35302826837979386552011-01-24T02:54:15.896+05:302011-01-24T02:54:15.896+05:30Just one question about this problem:
When calcul...Just one question about this problem:<br /><br />When calculating distance traveled along the horizontal direction shouldn't we also include friction which would also be time dependent?<br /><br />Force F also has a component along the vertical direction, so friction would be:<br /><br />(mg - at sin (alpha))kAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-5997147895880585179.post-37548419305285979562010-02-04T21:10:34.431+05:302010-02-04T21:10:34.431+05:30Thanks for the quick reply.
That was the key point...Thanks for the quick reply.<br />That was the key point I was missing. Should have figured it out since the problem says the force applied is a function of time and hence acceleration (F=ma) is a function of time too. <br />I suggest you include a one line explanation per the above in you solution for some of us who have this doubt.<br />Thanks Krishna. You have done a great job on this blog.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-5997147895880585179.post-50916826199435812882010-02-02T22:15:50.301+05:302010-02-02T22:15:50.301+05:30The equation v = u + acc*t is correct only when th...The equation v = u + acc*t is correct only when the acceleration acc is constant with time not when acc is changing with time. <br /><br />For example if acc = bt then,<br /> v = bt^/2 + ut. <br /><br />In this problem acc is changing with time.Krishna Kant Chintalapudihttps://www.blogger.com/profile/07024636005277277405noreply@blogger.comtag:blogger.com,1999:blog-5997147895880585179.post-85265312790265019912010-02-02T10:23:06.542+05:302010-02-02T10:23:06.542+05:30Why can't I say:
F cosx = m*acc
=>acc = Fco...Why can't I say:<br />F cosx = m*acc<br />=>acc = FcosX/m<br />=>acc = at*cosX/m<br /><br />Then v = u + at. u = 0<br />Hence v = (at * cosX/m) * t <br /><br />Substitute for t from equation 2.<br />It gives a different answer. Can you explain why ?Anonymousnoreply@blogger.com