Question

A block of 5kg is kept on a block of 10kg and the 10kg block is kept on a 20kg block. The friction coefficient between any two contacting surfaces is j = 0.5. A horizontal force is applied on 10kg blocks as shown. The maximum value of F(F+0) so that system moves together starting from rest is: g = 10 m/s2 5 10 20​

Answer 1

Answer:

10feet*10 kg

I hink this gives u answer

Answer 2

Answer:

Correct options are A) , B) , C) and D)

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.