The acceleration of the box is [tex]4.37\cdot 10^{-3} m/s^2[/tex]
Explanation:
The acceleration of the crate can be found by analyzing the forces acting along the direction parallel to the incline. We have three forces:
Taking into account the direction of each force, the equation of motion along this direction is:
[tex]F-F_f -W sin \theta = ma[/tex]
where
F = 125 N
[tex]F_f = 25.0 N[/tex]
[tex]W=292 N[/tex] is the weight of the crate
[tex]\theta=20.0^{\circ}[/tex] is the angle of the incline
[tex]m=\frac{W}{g}=\frac{292}{9.8}=29.8 kg[/tex] is the mass of the box
a is the acceleration
And solving for a, we find:
[tex]a=\frac{F-F_f - Wsin \theta}{m}=\frac{125-25-(292)(sin 20^{\circ})}{29.8}=4.37\cdot 10^{-3} m/s^2[/tex]
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