A 2.75-kg cat moves in a straight line (the x-axis). Figure E4.14 shows a graph of the x-component of this cat’s velocity as a function of time. (a) Find the maximum net force on this cat. When does this force occur? (b) When is the net force on the cat equal to zero? (c) What is the net force at time 8.5 s?

The mass of the cat is

$\text{m}=2.75\text{kg}$

The velocity of the cat is depicted by the given velocity-time graph.

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

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

The acceleration of a body is its rate of change in velocity. For constant acceleration

$\mathbf{a}\mathbf{=}\frac{{\mathbf{v}}_{\mathbf{2}}\mathbf{-}{\mathbf{v}}_{\mathbf{1}}}{{\mathbf{t}}_{\mathbf{2}}\mathbf{-}{\mathbf{t}}_{\mathbf{1}}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\left(2\right)$

Here, ${\mathrm{v}}_2$ is the velocity in time ${\mathrm{t}}_2$ and${\mathrm{v}}_1$ is the velocity in time t${\mathrm{t}}_1$ .

Since net force is directly proportional to acceleration and acceleration is the rate of change of velocity, the net force is maximum where the rate of change of velocity and hence the slope of the velocity-time graph is maximum. Inspection of the graph shows that the slope is maximum from 0 to 2 s. From equation (2), the acceleration in this time period is

$\begin{array}{rcl}\mathrm{a}& =& \frac{8\mathrm{m}/\mathrm{s}-0\mathrm{m}/\mathrm{s}}{2\text{s}-0\text{s}}\\ & =& 4\mathrm{m}/{\mathrm{s}}^2\end{array}$

From equation (1), the maximum net force is

$\begin{array}{rcl}\mathrm{F}& =& 2.75\text{kg}\times 4\mathrm{m}/{\mathrm{s}}^2\\ & =& 11·\left(1\text{kg}·\mathrm{m}/{\mathrm{s}}^2\times \frac{1\mathrm{N}}{1\text{kg}·\mathrm{m}/{\mathrm{s}}^2}\right)\\ & =& 11\mathrm{N}\end{array}$

Thus, the maximum net force applied on the crate is 11 N from 0 s to 2 s.

The net force is zero when the velocity of the cat is constant.

Inspection of the graph shows that the velocity is constant from 2 s to 6 s.

Thus, the net force is zero from 2 s to 6 s.

From equation (2), the acceleration of the cat between 6 s and 10 s is

$\begin{array}{rcl}\mathrm{a}& =& \frac{12\mathrm{m}/\mathrm{s}-8\mathrm{m}/\mathrm{s}}{10\text{s}-6\text{s}}\\ & =& 1\mathrm{m}/{\mathrm{s}}^2\end{array}$

From equation (1), the net force on the cat at 8.5 s is

$\begin{array}{rcl}\mathrm{F}& =& 2.75\text{kg}\times 1\mathrm{m}/{\mathrm{s}}^2\\ & =& 2.75·\left(1\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2\times \frac{1\mathrm{N}}{1\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2}\right)\\ & =& 2.75\mathrm{N}\end{array}$

Thus, the net force applied on the crate at 8.5 s is 2.75 N.