The system is released from
Rest & block B is found to
have a speed 0.3 m/s.
after it has descended
through a distance of 1m,
find the
coefficient of
friction b/w the block
and the table
(g=10m/s²)
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From constraint relations, we can see that
v
A
=2v
B
Therefore, v
A
=2(0.3)=0.6m/s
as v
B
=0.3m/s (given)
Applying W
net
=ΔU+ΔK
we get −μm
A
gS
A
=−m
B
gS
B
+
2
1
m
A
v
A
2
+
2
1
m
B
v
B
2
Here, S
A
=2S
B
=2 masS
B
=1 m (given)
∴−μ(4.0)(10)(2)=−(1)(10)(1)+
2
1
(4)(0.6)
2
+
2
1
(1)(0.3)
2
or −80μ=−10+0.72+0.045
or 80μ=9.235
or μ=0.115
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