Answer :
The acceleration of the box being pulled by the given force is 2.03 m/s².
The given parameters;
- mass of the box, m = 15 kg
- force applied to the box, F = 75 N
- angle of inclination of the force, θ = 20⁰
- coefficient of kinetic friction, μ = 0.33
The normal force on the box is calculated as follows;
[tex]F_n = mg- Fsin(\theta)\\\\F_n = 15\times 9.8\ - \ 75\times sin(20)\\\\F_n = 121.35 \ N[/tex]
The frictional force experienced by the box is calculated as follows;
[tex]F_k = \mu F_n\\\\F_k = 0.33 \times 121.35 \\\\F_k = 40 \ N[/tex]
The net horizontal force on the box is calculated as follows;
[tex]\Sigma F _x = 0\\\\Fcos(\theta) - F_k = ma\\\\[/tex]
where;
- a is the acceleration of the box
[tex]75 \times cos(20) \ - \ 40 = 15a\\\\30.425 = 15a\\\\a = \frac{30.425}{15} \\\\a = 2.03 \ m/s^2[/tex]
Thus, the acceleration of the box being pulled by the given force is 2.03 m/s².
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