Question

Two bodies A (400gm) and B are
connected by a light inextensible string
which passes over a frictionless pulley as
shown. If B comes down with acceleration
4 ms. Its mass is​

Answer 1

Given:-

• Mass of A = 400 gm = 0.4 kg
• Acceleration of block B = 4 m/s²
• The surface and the pulley is frictionless
• The string is light and inextensible

To find:-

• Mass of body B

Answer:-

As both the blocks A and B are connected by the same light inextensible string, acceleration of blocks A and B will be same,

$\sf \longrightarrow a_A = a_B = a = 4\: { }^{\sf m} \! / { }_{\sf{s}^{2}}$

Refer to the attachment for F.B.D.

For block A, the $\sf mg$ force has been resolved into its horizontal component $\sf mg\:cos(\theta)$ and $\sf mg\:sin(\theta)$ where $\sf \theta$ is the angle between $\sf mg$ and $\sf mg\:cos(\theta).$ And the horizontal and vertical components are perpendicular to each other.

For block A:-

$\sf T - \big\{m_A g\:sin(\theta)\big\} = m_A a$

Putting $\sf m_A = 0.4\:kg, \theta = 30^{o},$ and a = 4 m/s²,

$\longrightarrow \sf T - \big\{0.4\times 10 \times sin(30^o)\big\} = 0.4 \times 4$

$\longrightarrow \sf T - 2 = 1.6$

$\longrightarrow \sf T = 3.6\:N\quad\quad \dots (1)$

For block B:-

$\sf T - m_Bg = m_B(-a)$

$\longrightarrow \sf m_Bg - T = m_Ba$

$\longrightarrow \sf m_Bg - m_Ba = T$

$\longrightarrow \sf m_B(g - a) = T$

$\longrightarrow \sf m_B = \dfrac{T}{(g-a)}$

Putting $\sf T = 3.6\:N\; [From\:(1)],$ and a = 4 m/s²,

$\longrightarrow \sf m_B = \dfrac{3.6}{(10-4)}$

$\longrightarrow \sf m_B = \dfrac{3.6}{6}$

$\longrightarrow \sf m_B = 0.6\:kg$

$\longrightarrow \sf \underline{\underline{\sf{\green{ m_B = 600\:gm }}}}$