Question

a block is acted on by two forces as shown in the diagram below. if the magnitudes of the forces are $f_1 = 55.0$ n and $f_2 = 37.5$ n, what are the magnitude (in m/s²) and direction of the acceleration of the block? let $m = 8.00$ kg and $\theta = 38.0^{circ}$. oscolphys2016 4.7.wa.056.

Explanation:

Step1: Resolve $F_2$ into components

Resolve $F_2$ into horizontal and vertical components. The horizontal component of $F_2$ is $F_{2x}=F_2\cos\theta$ and the vertical component is $F_{2y} = F_2\sin\theta$.
$F_{2x}=37.5\cos(38.0^{\circ})\text{ N}\approx 29.5\text{ N}$

Step2: Calculate the net - force in the x - direction

The net - force in the x - direction, $F_{net,x}=F_1 + F_{2x}$.
$F_{net,x}=55.0\text{ N}+29.5\text{ N}=84.5\text{ N}$
The net - force in the y - direction, $F_{net,y}=F_{2y}$ (assuming no other vertical forces and no acceleration in the y - direction for a non - lifting situation). $F_{2y}=37.5\sin(38.0^{\circ})\text{ N}\approx 23.1\text{ N}$

Step3: Use Newton's second law to find the acceleration

According to Newton's second law $F = ma$. In the x - direction, $a_x=\frac{F_{net,x}}{m}$.
$a_x=\frac{84.5\text{ N}}{8.00\text{ kg}}\approx10.6\text{ m/s}^2$
Since there is no acceleration in the y - direction (assuming the block stays on the surface), the magnitude of the acceleration of the block is $a = a_x\approx10.6\text{ m/s}^2$ and the direction is along the positive x - direction.

Answer:

Magnitude: $10.6\text{ m/s}^2$, Direction: Along the positive x - direction (the direction of $\vec{F}_1$)