Question

a block of mass 3 kg which is on a smooth inclined plane making an angle of 30 degree to the horizontal is connected by cord passing over a light friction less Pulley to second block of mass 2 kg hanging vertically. what is the acceleration of each block and what is the tension of the cord ?​

Answer 1

Answer:

For the block that is hanging vertically, the equation will be;

20-T=2a.....(1)

For the block in inclined plane the equation will be;

T-3gsin30=3a.....(2)

From both equations ; a= 1m/s^2

And T=18N

Answer 2

Answer:

The acceleration of each block and tension of the cord are 3.92 m/s² and 17.64 N.

Explanation:

Given that,

Mass of first block = 2 kg

Mass of second block = 3 kg

Angle = 30°

We need to calculate the acceleration of each block

Using equation of balance

$m_{1}a=T-m_{1}g\sin\theta$....(I)

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

On adding equation (I) and (II)

$(m_{1}+m_{2})a=-m_{1}\times g\sin\theta+m_{2}g$

Put the value into the formula

$5a=2\times9.8\times\sin30+3\times9.8$

$a=\dfrac{2\times9.8\times\sin30+3\times9.8}{5}$

$a=7.84\ m/s^2$

The acceleration of each block is

$a=\dfrac{7.84}{2}$

$a=3.92\ m/s^2$

We need to calculate the tension

Using equation (II)

$2a=19.6-T$

Put the value in the equation

$m_{2}a=m_{2}g-T$

$-T=3\times3.92-3\times9.8$

$T=17.64\ N$

Hence, The acceleration of each block and tension of the cord are 3.92 m/s² and 17.64 N.