Two blocks are kept in contact with each other on a smooth surface. The force on the lighter block and its acceleration is:

6 N, 1 m / s^{2}

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4 N, 1 m / s^{2}

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10 N, 1 m / s^{2}

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0 N, 0 m / s^{2}

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Solution

The correct option is B $4 N, 1 m / s^{2}$
Given : $F = 10$ N $m_{1} = 6$ kg $m_{2} = 4$ kg
Total mass of the system $M = m_{1} + m_{2} = 6 + 4 = 10$ kg
$∴$ Net acceleration of the system $a = \frac{F}{M} = \frac{10}{10} = 1$ $m / s^{2}$
Thus acceleration of lighter block is $1$ $m / s^{2}$
Also force acting on the lighter block $F = m_{2} a = 4 \times 1 = 4$ $N$

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Two blocks are kept in contact with each other on a smooth surface. The force on the lighter block and its acceleration is:

The correct option is B 4N,1m/s2Given : F=10 N m1=6 kg m2=4 kgTotal mass of the system M=m1+m2=6+4=10 kg∴ Net acceleration of the system a=FM=1010=1 m/s2Thus acceleration of lighter block is 1 m/s2Also force acting on the lighter block F=m2a=4×1=4 N