A cylinder starting from rest slides down through the inclined plane of length 40 m with an acceleration of 20 m/sec what could be a speed at the bottom of the plane?
Answers
Answer:
Explanation:
When a plane is inclined to the horizontal at an angle θ, which is greater than the angle of repose, then the body placed on the inclined plane slides down with an acceleration a.
From figure,
R=mgcosθ ....(i)
Net force on the body down the inclined plane
f=mgsinθ−F ....(ii)
i.e., f=ma=mgsinθ−μR [∵μ=
R
F
]
∴ma=mgsinθ−μmgcosθ [from equation (i)]
=mg(sinθ−μcosθ)
⇒a=g(sinθ−μcosθ)
Time taken by the body to slide down the plane
t
1
=
a
2s
=
(gsinθ−μgcosθ)
2s
t
2
=
gsinθ
2s
[in absence of friction]
∵t
1
=2t
2
⇒t
1
2
=4t
2
2
[given]
∣
∣
∣
∣
∣
(gsinθ−μgcosθ)
2s
∣
∣
∣
∣
∣
2
=4×
∣
∣
∣
∣
∣
gsinθ
2s
∣
∣
∣
∣
∣
2
⇒
g(sinθ−μcosθ)
2s
=4×
gsinθ
2s
⇒sinθ=4sinθ−4μcosθ
⇒μ=
4
3
tanθ=
4
3
tan30
o
=0.43
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