Question

a cheetah has an acceleration that has x and y components of a_x = 5.3 m/s² and a_y = 3.8 m/s². the cheetahs mass is 61 kg. find the net force acting on the cheetah.
4.0×10² n, directed 54° above the x axis
5.6×10² n, directed 36° above the x axis
2.3×10² n, directed 54° above the x axis
3.2×10² n, directed 36° above the x axis
4.0×10² n, directed 36° above the x axis

Explanation:

Step1: Calculate x - component of force

According to Newton's second - law $F = ma$. The x - component of the force $F_x=ma_x$. Substitute $m = 61\ kg$ and $a_x = 5.3\ m/s^2$ into the formula: $F_x=61\times5.3 = 323.3\ N$.

Step2: Calculate y - component of force

The y - component of the force $F_y=ma_y$. Substitute $m = 61\ kg$ and $a_y = 3.8\ m/s^2$ into the formula: $F_y=61\times3.8 = 231.8\ N$.

Step3: Calculate the magnitude of the net force

The magnitude of the net force $F=\sqrt{F_x^{2}+F_y^{2}}$. Substitute $F_x = 323.3\ N$ and $F_y = 231.8\ N$ into the formula:
\[

$$\begin{align*} F&=\sqrt{(323.3)^{2}+(231.8)^{2}}\\ &=\sqrt{104522.89 + 53731.24}\\ &=\sqrt{158254.13}\\ &\approx397.8\ N\approx4.0\times 10^{2}\ N \end{align*}$$

\]

Step4: Calculate the direction of the net force

The direction $\theta$ of the net force is given by $\tan\theta=\frac{F_y}{F_x}$. Substitute $F_x = 323.3\ N$ and $F_y = 231.8\ N$ into the formula: $\tan\theta=\frac{231.8}{323.3}\approx0.717$. Then $\theta=\arctan(0.717)\approx36^{\circ}$ above the x - axis.

Answer:

E. $4.0\times 10^{2}\ N$, directed $36^{\circ}$ above the x - axis