Question

a construction worker pulls a (9.34\text{ kg}) toolbox at a constant speed, applying a force of (22.3\text{ n}) at an angle of (45.0^{circ}).

what is the coefficient of friction?

(mu_k = ?)

Explanation:

⚡ Using what you learned: newton's laws of motion

Step 1: Identify given values

$$ m = 9.34\text{ kg} $$

$$ F = 22.3\text{ N} $$

$$ \theta = 45.0^\circ $$

$$ g = 9.8\text{ m/s}^2 \quad (\text{or } 9.81\text{ m/s}^2) $$

Since the speed is constant, the acceleration is zero (\(a = 0\)), meaning the net force in both the horizontal and vertical directions is zero.

Step 2: Analyze horizontal forces

The horizontal component of the pulling force balances the kinetic friction force:

$$ F_x - f_k = 0 $$

$$ F \cos(\theta) - \mu_k F_N = 0 $$

$$ f_k = F \cos(\theta) = 22.3 \cdot \cos(45.0^\circ) \approx 15.768\text{ N} $$

Step 3: Analyze vertical forces

The vertical forces acting on the toolbox are the normal force upward, the vertical component of the pulling force upward, and gravity downward:

$$ F_N + F_y - F_g = 0 $$

$$ F_N + F \sin(\theta) - mg = 0 $$

$$ F_N = mg - F \sin(\theta) $$

$$ F_N = (9.34 \cdot 9.8) - 22.3 \cdot \sin(45.0^\circ) $$

$$ F_N \approx 91.532 - 15.768 = 75.764\text{ N} $$

(Using \(g = 9.81\text{ m/s}^2\): \(F_N \approx 91.625 - 15.768 = 75.857\text{ N}\))

Step 4: Calculate the coefficient of kinetic friction

$$ \mu_k = \frac{f_k}{F_N} $$

$$ \mu_k = \frac{15.768}{75.764} \approx 0.208 $$

(Using \(g = 9.81\text{ m/s}^2\): \(\mu_k = \frac{15.768}{75.857} \approx 0.208\))

Answer:

0.208