Two bodies having masses m1=40 g and m2=60 g are attached to the ends of a string of negligible mass and suspended from massless pulley. The acceleration of the bodies is:

Question

Two bodies having masses m1=40 g and m2=60 g are attached to the ends of a string of negligible mass and suspended from massless pulley. The acceleration of the bodies is: (A)  1 m/s2  (B)  2 m/s2  (C)  0.4 m/s2  (D)  4 m/s2

Tardigrade Admin · Accepted Answer

Let a be acceleration in the masses and

T

be the tension in the string. The equations of motion are

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

... (i)

T - m_{1} g = m_{1} a

... (ii)
From Eqs. (i) and (ii), we get

a = \frac{m _{2} - m _{1}}{m _{1} + m _{2}} g

Given,

m_{2} = 60 g, m_{1} = 40 g, g = 10 m / s^{2}

∴ a = \frac{60 - 40}{( 60 + 40 )} \times 10 = 2 m / s^{2}

Q. Two bodies having masses $m_{1} = 40 g$ and $m_{2} = 60 g$ are attached to the ends of a string of negligible mass and suspended from massless pulley. The acceleration of the bodies is:

$1 m / s^{2}$

$2 m / s^{2}$

$0.4 m / s^{2}$

$4 m / s^{2}$

Q. Two bodies having masses m1​=40g and m2​=60g are attached to the ends of a string of negligible mass and suspended from massless pulley. The acceleration of the bodies is:

1m/s2

2m/s2

0.4m/s2

4m/s2

Let a be acceleration in the masses and T be the tension in the string. The equations of motion are m2​=g−T=m2​a ... (i) T−m1​g=m1​a ... (ii) From Eqs. (i) and (ii), we get a=m1​+m2​m2​−m1​​g Given, m2​=60g,m1​=40g,g=10m/s2 ∴a=(60+40)60−40​×10=2m/s2

Q. Two bodies having masses $m_{1} = 40 g$ and $m_{2} = 60 g$ are attached to the ends of a string of negligible mass and suspended from massless pulley. The acceleration of the bodies is:

$1 m / s^{2}$

$2 m / s^{2}$

$0.4 m / s^{2}$

$4 m / s^{2}$