Three blocks are connected by massless strings on a frictionless inclined plane of 30° as shown in the figure. A force of 104 N is applied upward along the incline to mass m3 causing an upward motion of the blocks. What is the acceleration of the blocks? (Assume, acceleration due to gravity, .g=10 m / s 2)

Question

Three blocks are connected by massless strings on a frictionless inclined plane of 30° as shown in the figure. A force of 104 N is applied upward along the incline to mass m3 causing an upward motion of the blocks. What is the acceleration of the blocks? (Assume, acceleration due to gravity, .g=10 m / s 2) (A) 6.0 m / s 2 (B) 4.5 m / s 2 (C) 3.0 m / s 2 (D) 1.5 m / s 2

Tardigrade Admin · Accepted Answer

The given situation is shown in the figure below,

If the system moves upward with acceleration

am / s^{2}

on the application of

104 N

, then the equation of motion for

8 k g

block is

104 - T_{1} - 8 g sin 3 0^{\circ} = 8 a

\Rightarrow 104 - T_{1} - 8 \times 10 \times \frac{1}{2} = 8 a [∵ g = 10 m / s^{2}]

64 - T_{1} = 8 a \dots (i)

Equation of motion for

5 k g

block is

T_{1} - T_{2} - 5 g sin 3 0^{\circ} = 5 a

T_{1} - T_{2} - 25 = 5 a ... (ii)

Equation of motion for

3 k g

block is

T_{2} - 3 g sin 3 0^{\circ} = 3 a

T_{2} - 15 = 3 a ... (iii)

Adding Eqs. (i), (ii) and (iii), we get

24 = 16 a

\Rightarrow a = 1.5 m / s^{2}

Hence, acceleration

a

of block is

1.5 m / s^{2}

.

$6.0 m / s^{2}$

$4.5 m / s^{2}$

$3.0 m / s^{2}$

$1.5 m / s^{2}$

6.0m/s2

4.5m/s2

3.0m/s2

1.5m/s2

$6.0 m / s^{2}$

$4.5 m / s^{2}$

$3.0 m / s^{2}$

$1.5 m / s^{2}$