Question

FIGURE EX5.13 shows an acceleration-versus-force graph for a $500\text{g}$object. What acceleration values go in the blanks on the vertical scale?

Short answer

a.The acceleration of the object is$2.0\text{m}/{\text{s}}^2$.

Step-by-step solution

Part.a.

Newton's second law states that the acceleration acting on an object is directly proportional to the force acting on the object and inversely proportional to the mass of the object.

$a=\frac{F}{m}$

Here, $F$is the force acting on the object, and $m$is the mass of the object.

Step.2

The figure is shown below

From the figure at point$\left(\text{a}\right)$, the force is$2\text{N}$

Convert the unit of mass fromto

$\begin{array}{l}m=500\text{g}\left(\frac{{10}^{-3}\text{kg}}{1\text{g}}\right)\\ =500\times {10}^{-3}\text{kg}\end{array}$

Substitute $2\text{N}$for $F$, and $500\times {10}^{-3}\text{kg}$for $\text{m}$.

$\begin{array}{l}a=\frac{2\text{N}}{500\times {10}^{-3}\text{kg}}\\ =4.0\text{m}/{\text{s}}^2\end{array}$

Therefore, the acceleration of the object is$4.0\text{m}/{\text{s}}^2$.

Part.b.

From the figure at point (b), the force is$1\text{N}$

The acceleration acting on the object is,

$a=\frac{F}{m}$

Substitute $1\text{N}$for $F$, and $500\times {10}^{-3}\text{kg}$for $m$.

$\begin{array}{l}a=\frac{1\text{N}}{500\times {10}^{-3}\text{kg}}\\ =2.0\text{m}/{\text{s}}^2\end{array}$

Therefore, the acceleration of the object is$2.0\text{m}/{\text{s}}^2$.