91Ó°ÊÓ

$M_A=30\mathrm{kg},M_B=40\mathrm{kg},M_c=10\mathrm{kg}$

Newton’s second law of motion states that the net force acting on the body is equal to the product of mass and the acceleration of the body. The free body diagram gives us the different forces acting on the body along with their directions.

Using the Force body diagram and Newton’s second law we can find the tension in the cord and the displacement of A in first .

Newton’s second law is,$F_{net}=\underset{}{∑}Ma$

According to Newton’s second kinematic equation,

$∆x=\left(Vi\right)\left(t\right)+\frac{1}{2}a{t}^2$

Both, boxed B and C are moving down with an acceleration a. Assuming A and B as a one system, we can draw FBD for a complete system as below:

The total mass of the system is,

$M=40+10+30=80\mathrm{kg}$

From FBD we can write as,

$$\begin{gathered} M_{BC}×g=Ma \\ a=\frac{M_{BC}×g}{M} \\ =\frac{50\left(9.8\right)}{80} \\ =6.1\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

FBD for block C:

From FBD we can write as,

$$\begin{gathered} M_{c}g-T_{BC}=M_{C}a \\ {T}_{BC}=10\left[9.8-6.125\right] \\ =36.8\mathrm{N} \end{gathered}$$

The tension in the cord connecting B and C is 36.8N.

According to the newton’s second kinematic equation,

$∆x=\left(V_i\right)\left(t\right)+\frac{1}{2}a{t}^2$

So, the displacement of A in first 0.25 seconds is,

$$\begin{gathered} x-0=0+\left(\frac{1}{2}\right)\left(6.1\mathrm{m}/{\mathrm{s}}^2\right){\left(0.25\mathrm{s}\right)}^2 \\ =0.191\mathrm{m} \end{gathered}$$

Block A moves by a distance of 0.191m in first 0.25 s.