Q13E - University Physics with Modern Physics

Question

A 4.50-kg experimental cart undergoes an acceleration in a straight line (the x-axis). The graph in Fig. E4.13 shows this acceleration as a function of time. 

(a) Find the maximum net force on this cart. When does this maximum force occur? (b) During what times is the net force on the cart a constant? (c) When is the net force equal to zero?

Step-by-Step Solution

Answer

(a) The force applied on the 4.5 kg experimental card has a maximum value of 45 N between 2 s and 4 s.

(b) The force on the experimental cart is constant between 2 s and 4 s.

(c) The force on the experimental cart is zero at 0 s and 6 s.

     

Step 1: Given data

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.

Step 2: Second law of motion

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

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

Step 3: Maximum net force on the cart

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

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

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

Step 4: Constant net force

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.

Step 5: Zero net force on the crate

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.