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A block of mass M is kept in elevator (l...


A block of mass M is kept in elevator (lift) which starts moving upward with constant acceleration b as shown in figure. Initially elevator at rest. The block is observed by two observers A and B for a time interval `t=0` to `t=T`. Observer B is at rest with respect to elevator and observer A is standing on the ground.
Q. The observer A finds that the work done by gravity on the block is

Text Solution

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` ` `v=0+alpha1,,t_(1)(V)/(alpha)`
`S_(1)=(1)/(2)alphat_(1)^(2)=(v^(2))/(2alpha)`
O-V=`betat_(3),t_(3)=(v)/(beta)`
`O^(2)-V^(2)=-2|`
`S_(3)=(v_(2))/(2beta)=(1)/(2)betat_(3)^(2)`
`t_(2)=(S_(2))/(V)=(S-(S_(1)+S_(3)))/(V)=(S-(v^(2))/(2)((1)/(alpha)+(1)/(beta)))/(V)`
`t=t_(1)+t_(2)+t_(3)`
`=(v)/(alpha)+(s)/(v)-(v)/(2alpha)-(v)/(2beta)+(v)/(beta) t=(s)/(v)+(v)/(2)((alpha+beta)/(alphabeta))`
`S_(2)=S-(S_(1)+S_(2))=S-((v^(2))/(2alpha)+(v^(2))/(2alpha))`
`=S-(v^(2))/(alpha)`=s-(vt-s) `S_(2)`=2s-vt
[`S_(2)`=distance travelled with constant velocity]
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