IIT JEE , AIEEE Forum - Physics , Mechanics

michaelfaraday

Q. a block of 2kg is suspended from the ceiling through a massless spring of spring constant k = 100 N/m. a) what is the elongation in the spring? b) if another i kg is added to the block, find the further elongation?

then suppose if the ceiling is that of the elevator which is going up with an acceleration of 2m/s^2 find the elongations.

please explain the problem anyone......

sboosy

ganesha1991

a)
the elongation = x
at this point it is at rest
kx = mg
x = mg/k
= 2*10/100
= 1/5m
= 0.2m

michaelfaraday

ya but what about the second part....plz try that one...

ganesha1991

second case
its mi\oving up with a=2
now a pseudo force acts
therefore
kx = mg +ma
100x = 20 + 4
x = 24/100
= 0.24m

sboosy

Since the lift is accelerating up .mark a pseudo force down on the block = ma = 2m (in this case)

Thus in equilibrium we have

$mg + ma = kx \Rightarrow 12m = 100x \\ x = \frac{24}{100} = 0.24$

In case a one kg block is also added we have

$12m=100x \ \mbox{mass is 3 kg here} \\ x = \frac{36}{100} = 0.36 \ \mbox{m}$

Further elongation is 0.12 m

tarun_bits

see if you want to clear frames

http://www.goiit.com/posts/list/community-shelf-all-about-frames-the-fun-way-50388.htm

varun.tinkle

F=MG=KX
X=MG/K=20/100=0.2CM
WHEN THE BLOCK OF MASS 1KG IS ADDED THE TOTAL MASS BECOMES 3KG
EXTENSION=30/100=0.3
FURTHER ELINGATION=0.3-0.2=0.1
WHEN THE SPRING IS IN THE ELEVATOR THE GRAVUTY IS 8
SO THE FORCE IS
MG=3*8/100=0.24CM
PLZ RATE ME IF U FIND ME USEFUL
!!!!!!!!CHEERS!!!!

ganesha1991

hmmmm
varun.tinkle bro
u have done a mistake
the answer
is .24m
not .24cm

kalyan_iit

a]
the elongation = x
at this point it is at rest
kx = mg
x = mg/k
= 2*10/100
= 1/5m
= 0.2m

b]
second case
its moving up with a=2
now a pseudo force acts
therefore
kx = mg +ma
100x = 20 + 4
x = 24/100
= 0.24m

krishna.gopal

First case elongation is 2*g/k
Second case it is 3*g/k
Third case it will be 3*(g+a)/k
where a is the upward acceleration of lift

sarthak321

a)

let us assume that the spring is extended by elongation x.

when in eqillibrium state the net force on bock would be zero... refer to the fbd below....

forces on the body are :::

1. mg downwards....( 2 X 10 N in this case)...

2. Kx upwards ...... ( as spring exerts force in the opposite dirn of extension so as to .... restore its length..)

so... net force is 0 in equillibrium state....

mg - kx = 0 .....taking +ve in downwards dirn..

mg = kx

20 = 100 X x

x = 1/5 m

= 0.2 m

>>>>>>>>><<<<<<<<<

b) Forces in this case are ...

1. Mg downwards ....( note that M = ( 2 + 1)kg.... therfore Mg = 3 X 10 N)

2.kx' ( note x' is different from x )

applying the same concept as above and making net force 0

Mg = kx'

30 = 100 x'

x'= 0.3m

<<<<<<<<>>>>>>>

P.S ::::: Rate me if my answer helped....

sarthak321

Ans 2...

il not use the concept of pseudo force as it ill confuse u....

no if the the ceiling move up with acceleration then the mass m will aslo move up with the same acceleration i,e

2ms^2

now accelration is upwards ..... kx is upwards.....and mg is downwards....

net force upwards give acceleration upwards.

kx - mg = ma....( again +ve dirn is upwards...)

put the values ( mg = 20N .... k = 100N/m...... m = 2kg.....a = 2ms^2....) an get the answer as....

100x - 20 = 4

100x = 24

x = 0.24 m....

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