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Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class11  >>  Laws of Motion

A 5-kg block (m$_1$) and 2-kg block (m$_2$) are connected as shown on frictionless surface. Find the tension in the rope connecting the two blocks. Ignore any friction effects.

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Answer: a = 2.8 m/s$^2$, T = 14 N
We can draw the free body diagrams for our problem as follows:
When the system is released, let the acceleration of the blocks be $a$.
From the Free Body Diagrams, for the bigger block $m_1 \rightarrow T = m_1a$ and for the lighter block $m_2 \rightarrow m_2g - T = m_2a$
Solving for $a$, we get $a = \large\frac{m_2g}{m_1+m_2}$
Substituting for $a$ and solving for $T$, we get $T = m_1a = \large\frac{m_1m_2g}{m1+m_2}$
Given that $m_1 = 5 kg$ and $m_2 = 2 kg$:
$\Rightarrow a = \large\frac{2 \times 9.8}{5 + 2}$$ =2.8\; m/s^2$
Solving for $T$, we get $T = m_1a = 5 \times 2.8 = 14\;N$
answered Aug 19, 2014 by balaji.thirumalai
edited Aug 19, 2014 by balaji.thirumalai
 

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