Two masses m1 = 5 kg and m2 = 4.8 kg tied to a string are hanging over a light frictionless pulley. What is the acceleration of the masses when lift is free to move? (g = 9.8 ms-2)

Question

Two masses m1 = 5 kg and m2 = 4.8 kg tied to a string are hanging over a light frictionless pulley. What is the acceleration of the masses when lift is free to move? (g = 9.8 ms-2) (A) 0.2 ms-2 (B) 9.8 ms-2 (C) 5 ms-2 (D) 4.8 ms-2

Tardigrade Admin · Accepted Answer

On releasing, the motion of the sxstem will be according to figure

m_{1} g - T = m_{1} a ... (i)

and

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

On solving;

a = (\frac{m _{1} - m _{2}}{m _{1} + m _{2}}) g ... (iii)

Given,

m_{1} = 5 k g, m_{2} = 4.8 k g,

g = 9.8 m s^{- 2}

∴ a = (\frac{5 - 4.8}{5 + 4.8}) \times 9.8

\frac{0.2}{9.8} \times 9.8 = 0.2 m s^{- 2}

Q. Two masses $m_{1}$ = 5 kg and $m_{2}$ = 4.8 kg tied to a string are hanging over a light frictionless pulley. What is the acceleration of the masses when lift is free to move? (g = 9.8 $m s^{- 2}$ )

0.2 $m s^{- 2}$

9.8 $m s^{- 2}$

5 $m s^{- 2}$

4.8 $m s^{- 2}$

Q. Two masses m1​ = 5 kg and m2​ = 4.8 kg tied to a string are hanging over a light frictionless pulley. What is the acceleration of the masses when lift is free to move? (g = 9.8 ms−2)

0.2 ms−2

9.8 ms−2

5 ms−2

4.8 ms−2

On releasing, the motion of the sxstem will be according to figure m1​g−T=m1​a...(i) and T−m2​g=m2​a...(ii) On solving; a=(m1​+m2​m1​−m2​​)g...(iii) Given, m1​=5kg,m2​=4.8kg, g=9.8ms−2 ∴a=(5+4.85−4.8​)×9.8 9.80.2​×9.8=0.2ms−2

Q. Two masses $m_{1}$ = 5 kg and $m_{2}$ = 4.8 kg tied to a string are hanging over a light frictionless pulley. What is the acceleration of the masses when lift is free to move? (g = 9.8 $m s^{- 2}$ )

0.2 $m s^{- 2}$

9.8 $m s^{- 2}$

5 $m s^{- 2}$

4.8 $m s^{- 2}$