Two blocks of masses 40 kg and 30 kg are connected by a weightless string passing over a frictionless pulley as shown in the figure. The acceleration of the system would be

Question

Smart Achievers · Accepted Answer

(1)

Here, $m_{1} = 40 kg$

$m_{2} = 30 kg, θ = 30 °$

Let $T$ be the tension in the string and $a$ be the acceleration of the system.

Their equations of motion are

$m_{1} g \sin 30 ° - T = m_{1} a$ $\dots (i)$

$T - m_{2} g \sin 30 ° = m_{2} a$ $\dots (i i)$

Adding (i) and (ii), we get

$(m_{1} + m_{2}) a = (m_{1} - m_{2}) g \sin 30 °$

Substituting the given values, we get

$(40 + 30) a = (40 - 30) \times 9.8 \times \frac{1}{2} = 49$

$\Rightarrow$ $a = \frac{49}{70} = 0.7 {m s}^{- 2}$

Solution

Q.
Two blocks of masses 40 kg and 30 kg are connected by a weightless string passing over a frictionless pulley as shown in the figure. The acceleration of the system would be

1 0.7 $m s^{- 2}$

2 0.8 $m s^{- 2}$

3 0.6 $m s^{- 2}$

4 0.5 $m s^{- 2}$

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Solution

Q. Two blocks of masses 40 kg and 30 kg are connected by a weightless string passing over a frictionless pulley as shown in the figure. The acceleration of the system would be

1 0.7 m s-2

2 0.8 m s-2

3 0.6 m s-2