A 63.9kg water skier is pulled by a 125 N force at a 31.5 degrees angle, while the water creates a 84.5 N force pulling directly backward. What is the x-component of the acceleration?

Respuesta :

Answer:[tex]0.345 m/s^2[/tex]

Explanation:

Given

mass of skier [tex]m=63.9 kg [/tex]

Force [tex]F=125 N[/tex]

inclination of Force with horizontal [tex]\theta =31.5[/tex]

Water force [tex]F_w=84.5 N[/tex]

Resolving Forces

[tex]F\cos (31.5)-84.5=m\cdot a[/tex]

where a=acceleration of skier

[tex]125\cos (31.5)-84.5=63.9\times a[/tex]

[tex]a=\frac{106.58-84.5}{63.9}[/tex]

[tex]a=0.345 m/s^2[/tex]

This acceleration is in x direction therefore [tex]a_x=a=0.345 m/s^2[/tex]