Question

question 19 (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 component of the applied force? newtons apples 14.5 n 19° 4.72 n 14.5 n 4.99 n 13.7 n 5.41 n

Explanation:

Step1: Recall force - component formula

The formula for the horizontal component of a force $F$ at an angle $\theta$ with the horizontal is $F_x = F\cos\theta$.

Step2: Substitute values

Given $F = 14.5$ N and $\theta=19^{\circ}$, we have $F_x=14.5\times\cos(19^{\circ})$.
Using a calculator, $\cos(19^{\circ})\approx0.9455$, so $F_x = 14.5\times0.9455\approx13.7$ N.

Answer:

D. 13.7 N