The value of the acting force in the given diagram is
Acceleration = 0.1 ms-2
Mass = 10 kg
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Answer:
For 10kg block, f is frictional force
f
1max
=μ
1
μ
2
=μ
1
(10g)=0.1(10×10)=10N
i) if F=2N then acting friction
f
1
=F=2N
(A) is correct.
for 5 kg block;
f
2max
=μ(10+5)g=0.3×15×10=45N
ii) if F=30N then;
F−F
1max
=10a
10a=30−10=20
a=2m/s
2
(B) is correct.
iii) if μ
1
changed to 0.5;
then f
1max
=μ
1
(10g)=0.5×10×10=50N
In the case acting friction will be F
2max
Since,
F
1max
=F
2max
Block 10kg and 5kg both move together then F
min
to begin is
F
min
−F
max
=0
F
min
=F
2max
=45N
(c) is correct.
iv) if μ=0.1 , then f
max
=10N and,
f
2max
=45N
f
2max
>f
1max
always for any any value of N. Hence, 5kg block never move on the ground.
So, option (A), (B), (C), and (D) are correct.xplanation:
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