Question

Consider the situation shown in figure. Calculate (a) the acceleration of the $1.0kg$ blocks. (b) the tension in the string connecting the $1.0kg$ blocks and (c) the tension in the string attached to $0.50kg$.

Answer 1

From the free body diagram, we get

T + 0.5a - 0.5g = 0 -----------------(1)

$\mu R+1\times a+T_1-T$ = 0 -----------------(2)

$\mu R+1\times a-T_1$ = 0

$\mu R+a=T_1$ ------------(3)

From equations (i) , (ii) and (iii)

$\mu R+a=T-T_1$

$T-T_1=T_1$

T =$2T_1$

Equation (II) becomes $\mu R+a+T_1-2T_1$= 0

$\mu R+a-T_1$ = 0

$T_1=\mu R+a$

$T_1=0.2g+a$

Equation (i) will become

$2T_1+0.5a-0.5g=$ 0

$T_1=0.5g-\frac{0.5a}{2}$

= 0.25g - 0.25a

From equation (Iv) and (V)

0.2g +a = 0.25g - 0.25 a

a = $\frac{0.05}{1.25\times 10}$

a = 0.4$\times$ 10

a = 0.4 $m/s^2$

Therefore acceleration of 1kg block is 0.4 $m/s^2$

b) Tension $T_1=0.2g+a+0.4$ = 2.4 N

c) t = 0.5g - 0.5a

t = $0.5\times 10-0.5\times 0.4$

t = 4.8 N