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a block of mass 12 kg is released from rest om a friction-less incline of the angle theta=30 below the block is a spring having spring constant 10^4 N/m. the block stops momentarily when it compresses the spring by 4 cm. how far does the block move down the incline from its rest position to this stopping point
does anybody know
i asked this sol. many times but each time the FORMULA is different
irish boy come to physics department in my Q which you already answered but what youve answered is energy not the distance, please look other solutions too, i am getting diff. ans
there does seem to be some confusion about this. the first thing you need to do always is to draw it - and yes i could probably do this in my head but i would still always sketch it in outline just to be sure. that not only aids communications between you and others but will help you get the problem clear in your mind and it will also be a place to define variables such as vertical distance and distance along the slope - which are important here. you should do that first using the "Draw" widget that OS provides after that we can work through it.
yeah!! |dw:1434455454721:dw| this is what i mean by adding in other data. make sense?
yes it does
so "frictionless" means no energy lost to friction/heat, yes? that means we are looking at loss of grav potential energy of block which converts to potential energy in spring. simplest way to do it. is that what you are after?
yes that's what it mean and that's what i'm after
why not makes it easy for now just by calculating the energy in the spring when it is compressed 4cm. just work that out.
oka so this is the first step? 1/2 (10^4)(4)^2
mmmm. units? maybe this is where you are going wrong?
convert 4 cm into mterers so 0.04 m the ans i get K.E= 8J
i said that
8J here too all good so far now that is the grav potential energy the block lost by sliding down the slope. we need now calc h -- see the diagram. know how to?
U= mgh ? perhaps the P.E
i dont know, sighs, thats where i get confused perhaps we use energy conservation formula
looks good. do the sums and we can compare results. do you use g = 10 or 9.8?
9.8 to be exact
how to find h? where U= mgh U is??
we might resolve into its components ?
mgh sin theta
look at diagram where i marked "h" in. that is what we want. make sense? mgh = 8J
oh so U=K i see (12)(9.8)h= 8 so h is 0.06 m
check your calculator: 0.06 m ? i get 0.068 which is closer to 0.07.
yes yes it was
|dw:1434456513348:dw| so this is where we are, and another litle sketch to make sure we understand what is happening do you agree with this?
yes it right
and question asks: "how far does the block move down the incline from its rest position to this stopping point " do you see it? what is the answer?
mmm no, i dont. im sorry
the answer is h?
|dw:1434456741787:dw| to lose 8J of PE, the block had to fall 0.068m vertically. that involved travelling distance d down the slope. so this now becomes a simple trig problem....
do you know trigonometry? sin, cos, tan and the rest?
oh yes im sorry
you have the opposite, you want the hypoteneuse, sohcahtoa....
we're close, are you still there or wanna finish later?
im sorry there was a power cut
well, fire away. use trig and get d from h = 0.068 and angle = 30degrees
it would be dsin30=h so 0.07/sin30
is it correct
yes, in cm, d = 13.6cm
i saw this ans. somewhere else too, let me show you, please do visit