91Ó°ÊÓ

The mass of the cart is

$\text{m}=4.5\text{kg}$

The acceleration on the cart is depicted by the given acceleration-time graph.

The second law of motion relates the net force $F$, mass $m$ and acceleration $a$ as

$\mathbf{F}\mathbf{=}\mathbf{ma}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\left(1\right)$

From the figure, the maximum acceleration of the cart occurs between 2 s and 4 s and is equal to

$\text{a}=10\mathrm{m}/{\mathrm{s}}^2$

From equation (1), the maximum net force is

$\begin{array}{rcl}\mathrm{F}& =& 4.5\text{kg}×10\mathrm{m}/{\mathrm{s}}^2\\ & =& 45·\left(1\text{kg}·\mathrm{m}/{\mathrm{s}}^2×\frac{1\text{N}}{1\text{kg}·\mathrm{m}/{\mathrm{s}}^2}\right)\\ & =& 45\text{N}\end{array}$

Thus, the maximum net force applied on the crate is 45 N from 2 s to 4 s.

From the figure, the acceleration on the crate is constant from 2 s to 4 s. Thus, from equation (1), the net force on the crate is also constant from 2 s to 4 s.

From the figure, the acceleration on the crate is zero at 0 s and 5 s. Thus, from equation (1), the net force on the crate is also zero at 0 s to 6 s.