|
|
|
|
|

| Author |
Message |
![[Post New]](/templates/default/images/icon_minipost_new.gif) 25 Jun 2007 18:46:03 IST
|
|
|
a 1 kg block is situated on a rough incline is connected to a spring of spring constant 100 N/m.d block is released from rest wid d spring in d unstretched position. d block movs 10cm before coming to rest.find d coefficient of friction between d block and d incline.
ans) .125
|
we are the allies of konoha.....shinobis of the....sand
shubham sunder |
|
|
All India Online Test on 25th Jan 2009 for - IIT, AIEEE, KCET, EAMCET and many more .. |
| Enrol Free. Refer Friends. Win I-Phone 3G.
Register Now for FREE» |
|
|
![[Post New]](/templates/default/images/icon_minipost_new.gif) 25 Jun 2007 19:25:56 IST
|
|
|
what is the angle of inclination?
|
B.TECH SECOND YEAR
IIT DELHI
<TABLE CELLSPACING="1" CELLPADDING="1" BORDER="0">
<TR><TD>
<DIV ALIGN="right">Glitter Graphics</DIV></TD></TR></TABLE>
|
this reply: 0 points
(with 0 
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
All India Online Test on 25th Jan 2009 for - IIT, AIEEE, KCET, EAMCET and many more .. |
| Enrol Free. Refer Friends. Win I-Phone 3G.
Register Now for FREE» |
|
|
![[Post New]](/templates/default/images/icon_minipost_new.gif) 25 Jun 2007 19:34:11 IST
|
|
|
37degrees.....
sorry i forgot to mention it..
|
we are the allies of konoha.....shinobis of the....sand
shubham sunder |
this reply: 0 points
(with 0 
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
All India Online Test on 25th Jan 2009 for - IIT, AIEEE, KCET, EAMCET and many more .. |
| Enrol Free. Refer Friends. Win I-Phone 3G.
Register Now for FREE» |
|
|
![[Post New]](/templates/default/images/icon_minipost_new.gif) 25 Jun 2007 19:40:29 IST
|
|
|
i won't solve the problem because it's you who needs to solve it i can give you the guidelines after the block has come to rest the potential energy of the spring is increased and some energy has dissipated due to the friction force (which during the motion is kinetic friction which remains constant so this energy is force of friction X displacement) all this energy has come from the decrease in gravitational potential energy of the block. apply energy conservation and find the force of friction and hence the coefficient of friction.
|
B.TECH SECOND YEAR
IIT DELHI
<TABLE CELLSPACING="1" CELLPADDING="1" BORDER="0">
<TR><TD>
<DIV ALIGN="right">Glitter Graphics</DIV></TD></TR></TABLE>
|
this reply: 0 points
(with 0 
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
All India Online Test on 25th Jan 2009 for - IIT, AIEEE, KCET, EAMCET and many more .. |
| Enrol Free. Refer Friends. Win I-Phone 3G.
Register Now for FREE» |
|
|
![[Post New]](/templates/default/images/icon_minipost_new.gif) 25 Jun 2007 21:03:30 IST
|
|
|
Mass of the block, m = 1 kg Spring constant, k = 100 N/m Distance moved by the block, s = 10 cm = 0.1 m Angle of inclination, = 370
The block when released, attains a max strech for the spring, then it again goes towards the spring, during which it is stopped. When the block comes to rest, the force due to the spring is balanced by the frictional force and the component of mg.. kL = mgcos + mgsin L =( mgcos + mgsin )/k This is the strech of the spring in its equilibrium position. By energy considerations,
E = Wf
U + K = Wf Assuming the initial position of the block as the zero PE level, (-mgLsin - 0) + (0 - 0) = (- mgcos )s Lsin = scos [( mgcos + mgsin )/k]sin = scos
= sin /[(ks/mg)cot - cos ] Substituting the values
= 1.125
|
|
this reply: 0 points
(with 0 
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
All India Online Test on 25th Jan 2009 for - IIT, AIEEE, KCET, EAMCET and many more .. |
| Enrol Free. Refer Friends. Win I-Phone 3G.
Register Now for FREE» |
|
|
![[Post New]](/templates/default/images/icon_minipost_new.gif) 26 Jun 2007 22:39:56 IST
|
|
|
The block is attached to a spring fixed to the top of the inclined plane, right?? So it starts moving down. It stops and then comes back up. So the distance travelled by the block is greater than the final extension of the spring.
|
|
this reply: 0 points
(with 0 
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
All India Online Test on 25th Jan 2009 for - IIT, AIEEE, KCET, EAMCET and many more .. |
| Enrol Free. Refer Friends. Win I-Phone 3G.
Register Now for FREE» |
|
|
|
|
|
|
|
|
|