what is the acceleration along the ground of a 10 kg wagon when it is pulled with a force of 44 n at an…

Explanation:

Step1: Resolve the force horizontally

The horizontal - component of the force is given by $F_x = F\cos\theta$, where $F = 44\ N$ and $\theta=35^{\circ}$. $F_x = 44\cos35^{\circ}\approx44\times0.819 = 36.036\ N$

Step2: Use Newton's second - law

Newton's second - law is $F = ma$, where $F$ is the net force, $m$ is the mass, and $a$ is the acceleration. Here, the net force in the horizontal direction is $F_x$, and $m = 10\ kg$. $a=\frac{F_x}{m}$ Substitute $F_x = 36.036\ N$ and $m = 10\ kg$ into the formula: $a=\frac{36.036}{10}=3.6036\approx3.60\ m/s^{2}$

Answer:

$3.60\ m/s^{2}$