IIT JEE , AIEEE Forum - Physics , Mechanics

jhinuk_14

An elevator is descending with uniform accn,a person in the elevator drops a coin at the moment the elevator starts. the coin is 6 ft above the floor of the elevator at the time it is dropped. the person observes that the coin strikes the floor in one second. calculate frm these data the accn of the elevator.

elessar_iitkgp

We will work in the frame of reference of th lift.
Wrt the lift, the acceleration of the coin = a - g = a - 32 ft/s²
Dispacement of the coin wrt the lift = -h = -6 ft
Initial velocity of the coin wrt the lift = 0
Then,
-h = (1/2)(a-g)T²a = g - (2h/T²)
Substituting the values
a = 20 ft/s²

akku

Y did u take the accln as (a-g) and NOT (g-a) ?

PLSS REPLY

elessar_iitkgp

I am assuming +ve direction up. So the pseudo force on the elevator is ma up. Dur to this the acclen is +a. And due to gravity is -g

ramyani

I also liked this sum.
Look into the different solutions:

http://www.goiit.com/posts/list/mechanics-a-verma-sum-which-gave-me-pleasure-12157.htm

phyana

elessar is ur answer right.

the author plz give the right answer.

the coin is 6units above
the acc of the elevator and that of the coin is different.otterwise they wolud never meet.i assume coin acc=g
and elevator acc=g-a.

to get the accelaration equate the displacement

the n=ht of fall of coin =.5 * g * 1^2
= 5m=5*100/30 ft
but the displacement of the feet ot the elevator=500/30 - 6
this displacement=0.5 * (g-a) * 1^2

thus u get the acc=11.32 units(in ft)

plz check this elessar

jhinuk_14

i think elessar gave the rght ans. ....coz the bk says the ans. is 20 ft/sec^2

elessar_iitkgp

@phyana

The acceleration of the elevator isn't g-a. Its a.

You are trying to work in the reference frame of earth. So the equation will be
Displacement of the coin wrt earth = Disp of the coin wrt lift + displacement of the lift wrt earth
-(1/2)gT² = -h -(1/2)aT²
See, understand this, no matter which frame u work in, the equation will come out to be the same if the argument is sound

joyfrancis

Since the coin is undergoing free fall , the displacement in 1 sec is 4.9m
Displacement of the lift in 1 sec = a/2
since the coin is initially at 1.8m above the floor the net displacement of the coin is also equal to
1.8+a/2
Hence
1.8+a/2 = 4.9
a = 6.2m/s^2 = 20.5ft/s^2

omkarnet

answer is 20.34 ft/s^{2

phyana is wrongggggggggggg}

catch_arnnie

if g = 32 ft/sec. then ellessar gave the right ans.

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