Block a has a weight of 300 n and block b has a weight of 50 n. If the coefficient of kinetic friction between the incline and block a is . The speed, in ms1, of block a after it moves 1 m down the plane, starting from rest is , where * is not readable. Find *. Neglect the mass of the chord and pulleys. Take g = 10 ms2
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Answer:
N=300cosθ
=300×
5
4
=240N
Let us take,
Velocity of block be A after it moves 1 m
down the plane be V
and velocity of block B be V
B
We know that,
in pulley system
T.U = constant
⇒27.u=πu
B
⇒u
B
−2u
and T.x= constant
⇒27×1=f
2
x=2m
Since N=240N
⇒f=uN=0.2×240
f=48N
Applying energy conservation
Decrease in potential energy of block A
= Increase in kinetic energy of block A
+ Increase in kinetic energy of block B
+ Increase in potential energy of block B
+ work done by friction force
⇒300×1sinθ=
2
1
×
g
300
×v
2
+
2
1
×
g
50
(2v)
2
+50×2+48×1
300×
5
3
=15v
2
+10v
2
+100+48
=180=25v
2
+148
v
2
=
25
180−148
v
2
=
25
32
v=
5
4
2
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