Question

A 50.0-kg crate is being pulled along a horizontal smooth surface. The pulling force is 10.0 n and is directed 20.0° above the horizontal. What is the magnitude of the acceleration of the crate?

Answer 1

Answer:

$0.188 m/s^2$

Explanation:

Assuming the crate does not lift above the ground and remains along the floor, then its acceleration will be in the horizontal direction. Therefore, we can use Newton's second law to find its acceleration:

$F_x = ma_x$

where

$F_x$ is the net force on the crate along the x-direction

m is the mass of the crate

$a_x$ is the acceleration

Here we have:

m = 50.0 kg

$F_x = F cos \theta = (10.0 N)(cos 20.0^{\circ})=9.4 N$ is the component of the pulling force along the horizontal direction

Solving for the acceleration,

$a_x = \frac{F_x}{m}=\frac{9.4}{50.0}=0.188 m/s^2$