Question

Consider the situation shown in figure (6−E2). Calculate (a) the acceleration of the 1.0 kg blocks, (b) the tension in the string connecting the 1.0 kg blocks and (c) the tension in the string attached to 0.50 kg.

Figure

Answer 1

From the above diagrams:
T + ma − mg = 0
T + 0.5a − 0.5 g = 0 (1)
μR + ma + T₁ − T = 0
μR + 1a + T₁ − T = 0 (2)
μR + 1a − T₁ = 0
μR + a = T₁ (3)

From Equations (2) and (3) we have
μR + a = T − T₁
⇒ T − T₁ = T₁
⇒ T = 2T₁

So, Equation (2) becomes
μR + a + T₁ − 2T₁ = 0
⇒ μR + a − T₁ = 0
⇒ T₁ = μR + a
= 0.2g + a (4)

and Equation (1) becomes
2T₁ + 0.5a − 0.5g = 0

$$\begin{gathered} \Rightarrow T_1=\frac{0.5g-0.5a}{2} \\ =0.25g-0.25a\left(5\right) \end{gathered}$$

From Equations (4) and (5)
0.2g + a = 0.25g − 0.25a

$$\begin{gathered} \Rightarrow a=\frac{0.05}{1.25}\times 10 \\ =0.4\times 10m/s^2\left[g=10m/s^2\right] \end{gathered}$$

Therefore,
(a) the acceleration of each 1 kg block is 0.4 m/s²,
(b) the tension in the string connecting the 1 kg blocks is
T₁ = 0.2g + a + 0.4 = 2.4 N
​ and
(c) the tension in the string attached to the 0.5 kg block is
T = 0.5g − 0.5a
= 0.5 × 10 − 0.5 × 0.4
= 4.8 N.