Question

A 15.0 kg box is being pulled along level ground by a 75.0 N force that is directed at an angle of 200
above the horizontal. The coefficient of kinetic friction is 0.330. What is the acceleration of the box?

Answer 1

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;

$F_n = mg- Fsin(\theta)\\\\F_n = 15\times 9.8\ - \ 75\times sin(20)\\\\F_n = 121.35 \ N$

The frictional force experienced by the box is calculated as follows;

$F_k = \mu F_n\\\\F_k = 0.33 \times 121.35 \\\\F_k = 40 \ N$

The net horizontal force on the box is calculated as follows;

$\Sigma F _x = 0\\\\Fcos(\theta) - F_k = ma$

where;

• a is the acceleration of the box

$75 \times cos(20) \ - \ 40 = 15a\\\\30.425 = 15a\\\\a = \frac{30.425}{15} \\\\a = 2.03 \ m/s^2$

Thus, the acceleration of the box being pulled by the given force is 2.03 m/s².

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