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According to Newton's second law the net force $\left(F\right)$acting on an object is equal to the product of its mass$\left(m\right)$ and acceleration (a).

The expression for the net force acting on the object is as follows:

$F=ma$

The force acting on each of the masses remains the same. But, the mass and acceleration of the masses are different. Determine the force acting on different masses.

From the graph, the mass of the object is$100\text{g}$and the respective acceleration of the object is$15\text{m}/{\text{s}}^2$. The force on this object is calculated as follows:

Substitute $100\text{g}$for $m$and $15\text{m}/{\text{s}}^2$for $a$in the equation $F=ma$and calculate the force acting on $100\text{g}$object.

$\begin{array}{l}F=ma\\ =\left(100\text{g}\left(\frac{{10}^{鈭�3}\text{kg}}{1\text{g}}\right)\right)\left(15\text{m}/{\text{s}}^2\right)\\ =1.5\text{N}\end{array}$

Therefore, the force acting on this object is $1.5\text{N}$.

From the graph, the mass of the object is$200\text{g}$and the respective acceleration of the object is$7.5\text{m}/{\text{s}}^2$. The force on this object is calculated as follows:

Substitute $200\text{g}$for $m$and $7.5\text{m}/{\text{s}}^2$for $a$in the equation $F=ma$and calculate the force acting on $200\text{g}$object.

$\begin{array}{l}F=ma\\ =\left(200\text{g}\left(\frac{{10}^{鈭�3}\text{kg}}{1\text{g}}\right)\right)\left(7.5\text{m}/{\text{s}}^2\right)\\ =1.5\text{N}\end{array}$

Therefore, the force acting on this object is $1.5\text{N}$.

From the graph, the mass of the object is $300\text{g}$and the respective acceleration of the object is $5\text{m}/{\text{s}}^2$. The force on this object is calculated as follows:

Substitute $300\text{g}$for $m$and $5\text{m}/{\text{s}}^2$for $a$ in the equation $F=ma$ and calculate the force acting on $300\text{g}$object.

$\begin{array}{l}F=ma\\ =\left(300\text{g}\left(\frac{{10}^{鈭�3}\text{kg}}{1\text{g}}\right)\right)\left(5\text{m}/{\text{s}}^2\right)\\ =1.5\text{N}\end{array}$

Therefore, the force acting on this object is$1.5\text{N}$.

Hence, the force acting on all the three masses is same, and the force is $1.5\text{N}$.