Question

Find the acceleration of the 100 g block infig (819
1
Figure S-18
​

Answer 1

Answer:

Let us first calculate the limitation and kinetic friction between various surfaces. Between 3 kg and ground:

f

l

1

=f

k

1

=μ

1

(2+3)g=0.1×5g=5N

Between 2 kg and 3 kg:

f

l

2

=f

k

2

=μ

2

g=0.2×2g=4N

Let us first assume that both blocks move together with common accleration a.

a=

2+3

5+10−f

k

1

⇒a=

5

15−5

=2ms

2

Now let us see how much friction force is required between 2kg and 3kg for common acceleration a.

5−f

2

=2a⇒5−f

2

=2×2⇒f

2

=1N

Since f

2

<f

l2

,

1. Both blocks will move together with common accelerationa=2ms

−2

2. Friction between 2kg and 3 kg =f

2

= 1N Friction between 3 kg and ground = f

k

1

= 5N