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JEE Main Physics Class 11 Laws of Motion Part 2 Questions
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© examsnet.com
Question : 45
Total: 91
A lift of mass
M
=
500
kg
is descending with speed of
2
ms
−
1
. Its supporting cable begins to slip thus allowing it to fall with a constant acceleration of
2
ms
−
2
. The kinetic energy of the lift at the end of fall through to a distance of
6
m
will be _______
kJ
.
[31-Jan-2023 Shift 1]
Your Answer:
Validate
Solution:
v
2
=
u
2
+
2
as
=
2
2
+
2
(
2
)
(
6
)
=
4
+
24
=
28
KE
=
1
2
m
v
2
=
1
2
(
500
)
28
=
7000
J
=
7
kJ
Ans. 7
© examsnet.com
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