Question

question 20
4 points save answer
(problem reference 4 - 1) a 23.0 kg apple crate accelerates along a frictionless, horizontal surface as the crate is pulled with an applied force of 14.5 n as shown in the drawing.
what is the horizontal acceleration of the crate?
14.5 n
19°
newtons apples
a. 1.40 m/s²
b. 0.427 m/s²
c. 1.29 m/s²
d. 0.596 m/s²
e. 0.644 m/s²

Explanation:

Step1: Identify the horizontal - component of the force

We use the formula $F_x = F\cos\theta$, where $F = 14.5$ N and $\theta=19^{\circ}$. So $F_x=14.5\cos(19^{\circ})$.
$F_x = 14.5\times0.9455\approx13.71$ N

Step2: Apply 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. We know $m = 23.0$ kg and the net horizontal force $F_x\approx13.71$ N. Rearranging for $a$, we get $a=\frac{F_x}{m}$.
$a=\frac{13.71}{23.0}\approx0.596$ m/s²

Answer:

D. 0.596 m/s²