Question

a man drags a (10.0\text{ kg}) bag of mulch forward at a constant speed, applying a (22.5\text{ n}) force at (32.0^{circ}).

what is the friction force acting on the bag?

? n

remember to indicate if the force is positive (+) or negative (-).

Explanation:

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

Step 1: Analyze the forces in the horizontal direction

Since the bag moves forward at a constant speed, its acceleration is zero (\(a = 0\)). According to Newton's first law of motion, the net horizontal force must be zero:

$$ \sum F_x = 0 $$

The horizontal forces acting on the bag are:

• The forward horizontal component of the applied force: \(F_{app, x} = F_{app} \cos(\theta)\)
• The backward force of friction: \(F_f\)

$$ F_{app} \cos(\theta) + F_f = 0 $$

Step 2: Calculate the friction force

Given values:

• \(F_{app} = 22.5\text{ N}\)
• \(\theta = 32.0^\circ\)

Substitute these values into the equilibrium equation:

$$ 22.5 \cdot \cos(32.0^\circ) + F_f = 0 $$

$$ 22.5 \cdot 0.8480 + F_f = 0 $$

$$ 19.08 + F_f = 0 $$

$$ F_f = -19.1\text{ N} $$

Answer:

-19.1