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Two blocks of masses m1 = 5kg and m2 = 10kg are placed in the contact on the horizontal smooth surface as shown in the figure. A constant force of 150N applied horizontally on the block m, then the acceleration (in m/s2) of the system is :

 

 

 

Option: 1

10


Option: 2

15


Option: 3

30


Option: 4

35


Answers (1)

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Solution :

\\ \text{Given :} \\ m_{1}=5kg \\ m_{2} =10kg \\ F=150N,

\\ \text{Let acceleration of both the blocks be 'a'}

\\ \text{F.B.D of both blocks combined-}

\\ \text{From Newton's 2nd law of motion,}

F=(m_{1}+m_{2})a

\\ \therefore a = \frac{F}{m_{1}+m_{2}} = \frac{150}{15} = 10m/s^{2}

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Riya

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