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Question

One end of a spring of natural length 2m and spring constant k=100N/m is fixed at the ground and the other is fitted with a smooth ring of mass 1kg which is allowed to slide on a horizontal rod fixed at a height 2m (fig.) initially, then spring makes an angle of 37o with the vertical when the system is released from rest. Find the speed (in m/s) of the ring when the spring becomes vertical.
1078450_23c215e1991c454081d8c19c7f25629b.png

A
5m/s
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B
3m/s
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C
7m/s
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D
6m/s
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Solution

The correct option is A 5m/s

Given that,

Mass M=1kg

Natural length h=2m

Spring constant k=100N/m

Angle θ=370

Now,

cosθ=hl

l=hcosθ

l=20.798

l=2.50

Now, the extension

x=lh

x=2.502

x=0.5

Now, equating KE and PE

12mv2=12kx2

mv2=100×(0.5)2

mv2=100×.25

v2=25

v=5m/s

Hence, the speed of the ring is 5 m/s

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