Question

The situation shown in the figure , the coefficients of static friction and kinetic friction between mass m_1 and the horizontal surface are μ_s and μ_k , respectively. The blocks are released from rest. Which of the following statements are correct.

Note :- 1) Options and the diagram are in attachment.
2) Question contains multi answers.​

Answer 1

Answer:

Correct options are a), c), and d).

Explanation:

• The force of static friction acts on a body when it is at rest.
• The force of kinetic friction acts on a body when it is moving.
• The coefficient of static friction is always greater than the coefficient of kinetic friction for the same type of bodies in contact.

The formula used at the condition of equilibrium:

∑$F_{X}$ $=$ $0$

∑$F_{Y}$ $=$ $0$

where ∑$F_{X}$ is the sum of forces along $X$ axis

           ∑$F_{Y}$ is the sum of forces along $Y$ axis

Step 1:

At the condition of equilibrium:

$T=m_{1} g$

where $T$ is string tension

Also, $T=$ μs * $m_{2} g$

If blocks are at rest then,

$m_{1} g\leq$  μs * $m_{2} g$

$m_{1} \leq$  μs * $m_{2}$

So, the acceleration of blocks is zero and tension is  $m_{1} g$.

Step 2:

When the blocks are accelerating:

$m_{1}g >$  μs * $m_{2} g$

$m_{1} >$  μs * $m_{2}$

The friction on the block $m_{2}$ is μk * $m_{2} g$.

Step 3:

To calculate acceleration:

$m_{1} g-T=m_{1} a$

$T-$ μk$m_{2} g=m_{2} a$

So,  $m_{1} g-m_{2}a-$ μk$m_{2} g= m_{1} a$

$a=\frac{(m_{1}-u_{k} m_{2})g }{m_{1}+ m_{2} }$

So, acceleration is  $a=\frac{(m_{1}-u_{k} m_{2})g }{m_{1}+ m_{2} }$.

So, the correct options are a), c), and d) .