Question

The value of the acting force in the given diagram is
Acceleration = 0.1 ms-2
Mass = 10 kg​

Answer 1

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: