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Question


An ideal massless spring can be compressed 1 m by a force of 100 N. This same spring is placed at the bottom of a frictionless inclined plane which makes an angle
θ=30 with the horizontal. A 10 kg mass is released from rest at the top of the incline and is brought to rest momentarily after compressing the spring 2 meters.
(a) Through what distance does the mass slide before coming to rest?
(b) What is the speed of the mass just before it reaches the spring?

A
2m,v=25 m/s
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B
5m,v=25 m/s
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C
5m,v=5 m/s
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D
10m,5m,v=350 m/s
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Solution

The correct option is A 2m,v=25 m/s

Let total distance moved by the block is.
S=(l+2)m Where l is the distance moved by the block before touching the spring.
Now, work done by gravity on the block is Wg=mgSsinθ=10×10×Ssin30J
Wg=50SJ (1)
Work done by spring on the block is Ws=12kx2 Here K=100 N/m and x=2m
Ws=12×100×4J=Ws=200J (2)
Total work done W=Wg+Ws
W=(50 s - 200) J
Since change in K.E. of the block is zero as W=ΔK.E.
50s200=0 S=4m
(b) as S = l + 2
l = S - 2 = 2m
Work done by gravity over this path length is
Wg=mg×2msinθ=10×10×2×12=100J Wg=ΔK.E.
100=12mv20
v2=100×210=20
v=25 m/s

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