Question

A boy is standing on a weighing machine placed on the floor of an elevator.The elevator starts going down with some acceleration ,moves with constant velocity for some time and decelerates to stop.The minimum and maximum weight recorded are 480 and 840 N.If the magnitude is double of acceleration ,find (a)true weight of the boy and (b) magnitude of acceleration.


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Answer 1

Question:-

A boy is standing on a weighing machine placed on the floor of an elevator. The elevator starts going down with some acceleration, moves with constant velocity for some time and decelerates to stop. The minimum and maximum weight recorded are 480 and 840 N. If the magnitude of retardation is double of acceleration, find,

(a) true weight of the boy and

(b) magnitude of acceleration.

Answer:-

$\Large\boxed{\begin{minipage}{3cm}\mathsf{(a)\ 600\ N}\\\\\text{ \mathsf{(b)\ 2\ m\ s^{-2}} }\end{minipage}}$

Solution:-

Consider when the lift accelerates down. The weight of the boy is 'mg' downwards, apparent weight (reaction) is 'R' upwards and net force is 'ma' downwards. So,

$\longrightarrow\sf{mg-R=ma}$

$\longrightarrow\sf{R=m(g-a)}$

So there's loss of weight in the boy, and thus this is the minimum weight of the boy. Hence,

$\longrightarrow\sf{m(g-a)=480\ N}$

$\longrightarrow\sf{ma=mg-480\quad\quad\dots(1)}$

Consider when the lift decelerates to stop. The weight of the boy is also 'mg' downwards, apparent weight (reaction) is ' R' ' downwards and net force is '2ma' upwards, since magnitude of retardation is double of the acceleration.. So,

$\longrightarrow\sf{R'-mg=2ma}$

$\longrightarrow\sf{R=m(g+2a)}$

So there's gain of weight in the boy, and thus this is the maximum weight of the boy. Hence,

$\longrightarrow\sf{m(g+2a)=840\ N}$

$\longrightarrow\sf{mg+2ma=840}$

From (1),

$\longrightarrow\sf{mg+2(mg-480)=840}$

$\longrightarrow\sf{3mg-960=840}$

Then, weight of the boy,

$\longrightarrow\sf{\underline{\underline{mg=600\ N}}}$

Taking $\sf{g=10\ m\ s^{-2},}$

$\longrightarrow\sf{m=60\ kg}$

Then, again from (1), acceleration,

$\longrightarrow\sf{a=\dfrac{mg-480}{m}}$

$\longrightarrow\sf{a=\dfrac{600-480}{60}}$

$\longrightarrow\sf{\underline{\underline{a=2\ m\ s^{-2}}}}$

Answer 2

Answer:

here is ur solution

Explanation:

hope it helps