Physics, asked by harsh5544, 4 months ago

wyl
IA
B
In the arrangement shown in figure ma = 4kg
and mb = 1 kg. The system is released from
rest and block B is found to have a speed 0.3
m/s after it has descended through a
distance of 1 m. Find the coefficient of
friction between the block A and the table.
Neglect friction elsewhere.
(Take g = 10 m/s2):-​

Answers

Answered by zebaalmaas786
0

Answer:

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

Similar questions