Civil Engineering - Applied Mechanics - Discussion
Discussion Forum : Applied Mechanics - Section 3 (Q.No. 47)
47.
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
Discussion:
3 comments Page 1 of 1.
SHUBHAM SIDDHARTH NAYAK said:
5 years ago
Initail 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.
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.
(3)
Akki said:
5 years ago
Please someone give an explanation.
Dave said:
7 years ago
How, Can someone give me the detail?
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