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Applied Mechanics and Graphic Statics
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A 49 kg lady stands on a spring scale in an elevator. During the first 5 sec, starting from rest, the scale reads 69 kg. The velocity of the elevator will be

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Correct Answer: Option C
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4 Comments
Himanshu Das
Himanshu Das 2 years ago
To find a first
Since,weight increases moving upward,W=m(g+a)
49(9.81+a)=69*9.81
a=4m/s2
V=0+4*5
=20 m/s2
Kamal Jaswal
Kamal Jaswal 5 years ago
initial weight before moving on lift(m) =49 kg.

We know as we move upwards increase in weight occurs i.e acceleration increase by (g+a).
New weight (m1) = 69 kg.
Time(t) =5 sec.
Weight(W)= m(g+a) =m1 * g.
49(9.81+a) =69*9.81.
a=4m/sec2.
We know v=u+at.
u =0(as starts from rest).
v = 4*5=20 m /sec2.
Sabireen Sabireen
Sabireen Sabireen 5 years ago
the correct answer is [M-1 L3 T-2]
Patel Nisharg
Patel Nisharg 5 years ago
Asa ka kyo heee? Kyo ke +5 to app ne keya hee nai hiiiii