Question

A particle is projected at 60’ to the horizontal floor with a speed of 4m/s .it strike an uncompressed spring and compress it till the block is motionless .the fk=15N and k=10,000N/m,the spring compressed by..

Answer 1

a particle thrown vertically moves such that it passes from same height at 2 second and 10 seconds

Answer 2

a particle is released from a height H.at certain height it's k.e is three times it's P.E.find height and p.e at that instant

Answer 3

@Ruchi. Please post only one question at a time.
Thanks

Answer 4

u always suggest people nt reply of any 1 question

Answer 5

ha ha. I did reply yesterday but you logged off

Answer 6

Let's take this one
a 120m long train moves with speed 20m/s another train 130m long is moving with 30m/s in opposite direction.find much time they will cross?

Answer 7

@ash2326 i know that i hav to post 1 question at a time okay now reply at that i don't go offline bt after opening os my network work slowly i don't know why.

Answer 8

Are you using the latest version of chrome?

Answer 9

@ash2326 don't reply of first 2nd i hav solve it reply of 1st .if u know.

Answer 10

yes

Answer 11

hlp me out @ash2326

Answer 12

@apoorvk hlp me out

Answer 13

@lalaly

Answer 14

When does it strike the spring? I mean how far is the spring, does it strike completely horizontally, or obliquely?

Answer 15

from her diagram, i think we are supposed to assume it strikes horizontally.

Answer 16

i got \[F = mg/\mu _{k}\]........(i)
for the spring, F = kx
we'll know F from (i) and k is given. so we can find x, right?

Answer 17

so x = 0.025 m ?

Answer 18

Sorry I was away

Answer 19

i hav drawing diagram of another question

Answer 20

1st question is another one

Answer 21

Well, okay!
Assuming that, the block strikes horizontally,
it's velocity then = vcos60 = 4 (1/2) = 2 m/sec
K. Energy of the block at the time of collision = (1/2) x m x (v)^2 = 2m J

But! We cannot really use the principle of conservation of energy, since there is a non-conservative force (aka friction) acting in here. However if we can assume that the block remains horizontal in it's motion through out the displacement, then no friction would act on the block, and we can safely use this principle.

Either this, or am comprehending the scenario wrongly.

Answer 22

As for the force principle, I doubt we can use it since it's quite a dynamic system.

Answer 23

Let the compression be x
It's easy the kinetic energy of the block is transferred to the spring at potential energy and lost as heat due to friction
Velocity in horizontal direction= v cos 60=2 m/s
KE of the block as it strikes the spring= \(\frac 12m\times v^2\)
Energy lost due to friction=Fk *x
PE in Spring=\( \frac1 2 K\times x^2\)
so
\[\frac {1} 2 m\times v^2=\frac {1} 2 k\times x^2+ F_k x\]

Answer 24

@apoorvk what do you think?

Answer 25

i am nt getting u apporvk can u solve 2nd 1in which i hav drawn diagram

Answer 26

we're solving the originally posted question right?

Answer 27

Yeah

Answer 28

@Ruchi. what do you think?

Answer 29

what?

Answer 30

About my approach, see my previous post

Answer 31

@ash2326 I agree.

Answer 32

@Vaidehi09 @Ruchi. How do we find the mass?

Answer 33

Is it given ? @Ruchi.

Answer 34

\[\left[ a *b \right]=\sqrt{3}a .b\]

Answer 35

f=ma

Answer 36

find angle b/w a&b

Answer 37

where did a and b come from?

Answer 38

@Ruchi. We are discussing about the first question and you are posting a new question in between. This will only create confusion. !!

Answer 39

its vector question

Answer 40

okay

Answer 41

i hav nt much time that's why.

Answer 42

If you post several questions in one post. It won't help much.
No other user will come and help !

Answer 43

@ruchi. he's right. i already mixed up 2 sums!

Answer 44

@ash2326 the \(F_k\) would depend on the normal reaction from the spring (or whatever surface is attached to the spring) - and that is very dynamic, how do we find that?