The force that the 4.0 kg block exerts on the 5.0 kg block is 7.581N
According to Newton's second law of motion:
[tex]\sum F = ma[/tex]
If we have blocks with masses of 3.0 kg, 4.0 kg, and 5.0 kg are lined up in a row on a frictionless table, the total mass is expressed as:
Total mass = 3.0 + 4.0 + 5.0
Total mass = 12.0kg
Get the total acceleration "a"
[tex]a=\frac{\sum F}{m}\\a = \frac{13}{12.0} \\a=1.083m/s^2[/tex]
To calculate the force that the 4.0kg block exerts on the 5.0 kg block, we will use the formula above to have;
[tex]F = ma\\ F =(3.0+4.0)(1.083)\\F=7 \times 1.083\\F= 7.581N[/tex]
Hence the force that the 4.0 kg block exerts on the 5.0 kg block is 7.581N
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