Two masses of 10 kg and 5 kg are suspended from a rigid support as shown in figure. The system is pulled down with a force of 150 N attached to the lower mass. The string attached to the support breaks and the system accelerates downwards. If the downward force continues to act, the acceleration of the system is

Question

Two masses of 10 kg and 5 kg are suspended from a rigid support as shown in figure. The system is pulled down with a force of 150 N attached to the lower mass. The string attached to the support breaks and the system accelerates downwards. If the downward force continues to act, the acceleration of the system is (A) 20 ms-1 (B) 10 ms-1 (C) 5 ms-2 (D) zero

Tardigrade Admin · Accepted Answer

Given, Two masses 10 kg , 5 kg, F orce =150 N So the acceleration of the system is a=(F text net /m1+m2) ⇒ (F+(m1+m2) g/m1+m2) =(150+(10+5) × 10/10+5)=(300/15)=20 m / s 2

$20 m s^{- 1}$

$10 m s^{- 1}$

$5 m s^{- 2}$

$zero$

Given,
Two masses $10 k g, 5 k g$ ,
F orce $= 150 N$
So the acceleration of the system is $a = \frac{F _{net}}{m _{1} + m _{2}} \Rightarrow \frac{F + ( m _{1} + m _{2} ) g}{m _{1} + m _{2}}$
$= \frac{150 + ( 10 + 5 ) \times 10}{10 + 5} = \frac{300}{15} = 20 m / s^{2}$

20ms−1

10ms−1

5ms−2

zero

Given, Two masses 10kg,5kg, F orce =150N So the acceleration of the system is a=m1​+m2​Fnet ​​⇒m1​+m2​F+(m1​+m2​)g​ =10+5150+(10+5)×10​=15300​=20m/s2

$20 m s^{- 1}$

$10 m s^{- 1}$

$5 m s^{- 2}$

$zero$

Given,
Two masses $10 k g, 5 k g$ ,
F orce $= 150 N$
So the acceleration of the system is $a = \frac{F _{net}}{m _{1} + m _{2}} \Rightarrow \frac{F + ( m _{1} + m _{2} ) g}{m _{1} + m _{2}}$
$= \frac{150 + ( 10 + 5 ) \times 10}{10 + 5} = \frac{300}{15} = 20 m / s^{2}$