Two bodies of masses $4 k g$ and $3 k g$ respectively are connected by a light string passing over a smooth friction pulley. Calculate the acceleration of the masses and the tension in the string.

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Solution

From above image
for $4 k g$ block
$4 g - T = 4 a$ .. ..(1)
and for $3 k g$ block
$T - 3 g = 3 a$ . . .(2)
solving both equation we will get
acceleration = $\frac{g}{7} = \frac{10}{7}$ ${m / s}^{2}$
and tension = $\frac{24 g}{7} = \frac{240}{7}$ ${m / s}^{2}$

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Two bodies of masses 4 kg and 3 kg respectively are connected by a light string passing over a smooth friction pulley. Calculate the acceleration of the masses and the tension in the string.

From above image for 4kg block 4g−T=4a .. ..(1)and for 3kg blockT−3g=3a . . .(2)solving both equation we will get acceleration = g7=107 m/s2and tension = 24g7=2407 m/s2

Two bodies of masses $4 k g$ and $3 k g$ respectively are connected by a light string passing over a smooth friction pulley. Calculate the acceleration of the masses and the tension in the string.

From above image
for $4 k g$ block
$4 g - T = 4 a$ .. ..(1)
and for $3 k g$ block
$T - 3 g = 3 a$ . . .(2)
solving both equation we will get
acceleration = $\frac{g}{7} = \frac{10}{7}$ ${m / s}^{2}$
and tension = $\frac{24 g}{7} = \frac{240}{7}$ ${m / s}^{2}$