Question

When the three blocks of masses M, 2M and
2M in the system are released from rest,
they accelerate with a magnitude of 1 m/s2
What is the coefficient of kinetic friction
between block on the table and the table
top? (neglect the force of friction in the
pulleys and take g = 10 m/s2)​

Answer 1

Answer:

"is" (and any subsequent words) was ignored because we limit queries to 32 words.

Answer 2

Given:

The three blocks of masses M, 2M and 2M in the system are released from rest, they accelerate with a magnitude of 1 m/s².

To find:

Coefficient of friction between table and block2 ?

Calculation:

Let the tension between block 1 and block 2 be T1, and block 2 and block 3 be T2 :

For block 1 (M) :

$\therefore \: T_{1} - Mg = Ma \: \: \: .....(1)$

For block 2 (2M):

$\therefore \: T_{2} - T_{1} - \mu(2M)g = (2M)a \: \: \: ....(2)$

For block 3 (2M):

$\therefore \: (2M)g -T_{2} = (2M)a \: \: \: ....(3)$

Adding the equations:

$\implies \: 2Mg - Mg - 2 \mu Mg = Ma +2 Ma + 2Ma$

$\implies \: Mg - 2 \mu Mg = 5Ma$

$\implies \: M(10) - 2 \mu Mg = 5M(1)$

$\implies \: 10M- 20 \mu M= 5M$

$\implies \: 5M = 20 \mu M$

$\implies \: 5 = 20 \mu$

$\implies \: \mu = \dfrac{5}{20}$

$\implies \: \mu = 0.4$

So, coefficient of friction between table and block2 is 0.4 .

Hope It Helps.