Question

a cheetah has an acceleration that has x and y components of $a_x = 5.3 m/s^2$ and $a_y = 3.8 m/s^2$. the cheetahs mass is 61 kg. find the net force acting on the cheetah.
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 54° above the x - axis
4.0×10² n, directed 36° above the x - axis
5.6×10² n, directed 36° above the x - axis

Explanation:

Step1: Calculate the x - component of the net force

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

Step2: Calculate the y - component of the net force

Using Newton's second law again, the y - component of the force $F_y=ma_y$, with $m = 61\ kg$ and $a_y = 3.8\ m/s^2$. So $F_y=61\times3.8=231.8\ N$.

Step3: Calculate the magnitude of the net force

The magnitude of the net force $F=\sqrt{F_x^{2}+F_y^{2}}=\sqrt{(323.3)^{2}+(231.8)^{2}}=\sqrt{104522.89 + 53731.24}=\sqrt{158254.13}\approx397.81\ N\approx4.0\times 10^{2}\ N$.

Step4: Calculate the direction of the net force

The direction $\theta=\tan^{- 1}(\frac{F_y}{F_x})=\tan^{-1}(\frac{231.8}{323.3})\approx36^{\circ}$ above the x - axis.

Answer:

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