CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

An ideal massless spring can be compressed by 1m by a force of 100N. This same spring is placed at the bottom of a frictionless inclined plane which makes an angle θ=30o with the horizontal. A 10kg 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?
1018300_b47c38fde2cf412ca20c5b987b34c5aa.png

Open in App
Solution

(a) 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×Ssin30JWg=50SJ
Work done by spring on the block is, WS=12kx2WS=12×100×4J=WS=200J....(2)
Total work done W=Wg+WSW=(50S200)J
Since change in KE of the block is zero as W=ΔK.E50S200=0S=4m

(b) As S=l+2l=S2=2m

Work done by gravity over this path length is
Wg=mg×2sinθ=10×10×2×12=100J[Wg=ΔK.E]

100=12mv20=v2=100×210=20v=25m/s

1046374_1018300_ans_48933676260040cbb009d9a5a27996f6.png

flag
Suggest Corrections
thumbs-up
1
similar_icon
Similar questions
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
The Law of Conservation of Mechanical Energy
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon