Q14E - University Physics with Modern Physics

Question

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?

Step-by-Step Solution

Answer

(a) The maximum net force on the 2.75 kg cat is 11 N which is applied from 0 s to 2 s.

(b) The net force on the cat is equal to zero between 2 s and 6 s.

(c) The net force on the cat at time 8.5 s is 2.75 N.

Step 1: Given data

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.

Step 2: Second law of motion

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${\mathrm{t}}_1$ .

Step 3: Maximum net force on the cat

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.

Step 4: Zero net force on the cat

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.

Step 5: Force on the cat at 8.5 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.