Question

A block of mass m1 = 2.00 kg on smooth inclined plane at 30°is connected to second block of mass m2=3kg by cord passing over frictionless pulley .the acceleration of each block

Answer 1

draw the free body diagram

3g - T = 3a---- eq1

T - 2gsinθ= 2a --- eq2

by adding both equations

3g -- 2 x 10 x 1/2= 5a (sin 30°=1/2)

30-- 10 = 5a

a=4m/s2

Answer 2

The acceleration of each block would be $4 m/sec^2$

Explanation:

Given that,

$m_{1} = 2 kg$

The angle of inclined plane = 30°

$m_{2} = 3 kg$

Assuming T be the tension among the blocks and a be the acceleration, the two equations for the two blocks would be:

$m_{1} a = T - m_{1}gSin30$°  ...(i)

$m_{2} a = m_{2}g - T$            ...(ii)

After solving the two, we get,

T = $m_{1}(gSin30 + a)$   ...(i)

T = $m_{2} (g - a)$             ...(ii)

By solving these two,

$m_{1}(gSin30 + a)$ =  $m_{2} (g - a)$

⇒ a = $(m_{2} g - m_{1}gSin30)/(m_{1} + m_{2})$

⇒ a = (3 × 9.8 - 2 × 9.8 × 1/2)/(2 + 3)

⇒ a = 3.92 or approximately 4 m/sec^2

∵ a = $4 m/sec^2$

Learn more: Acceleration

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