Instructions Complete all work independently using correct physics notation, symbols, and SI units. Show all work using the GRASS problem-solving format (Given, Required, Analysis, Solution, Statement). Draw a fully labeled Free-Body Diagram (FBD) by hand. Record the required audio response using appropriate physics terminology. AI-generated solutions or other unauthorized assistance are not permitted. Problem: A 12.0 kg crate is pulled across a horizontal warehouse floor by a rope that makes an angle of 30.0° above the horizontal. The coefficient of kinetic friction between the crate and the floor is 0.25. The crate starts from rest and travels 18.0 m in 5.00 s with constant acceleration. a) Draw a fully labeled Free-Body Diagram (FBD) for the crate. Include all forces acting on the crate. b) Determine the acceleration of the crate using kinematics. c) Calculate the magnitude of the pulling force applied to the rope. Show all steps and include appropriate units. d) Record a 1–2-minute audio response explaining why the normal force is not equal to the weight of the crate in this situation, and describe how this affects the frictional force. Use physics terminology and refer to your FBD in your explanation. Rubric Category Criteria Marks G – Identifies relevant known quantities with correct symbols and units. 1 R – Clearly identifies what must be determined with correct symbols and units. 1 A – Includes a complete and correctly labeled FBD. Selects appropriate equations, resolves forces correctly, develops force relationships, and provides a clear audio explanation using proper physics terminology. 4 S – Performs calculations correctly, shows substitutions and algebraic steps, and uses appropriate units. 3 S – States the final answer clearly with correct units and significant digits. 1

The crate is not accelerating vertically, so $\Sigma F_y = 0$.
$F_N + F_{Ty} - F_g = 0$
$F_N + F_T \sin\theta - mg = 0$
From this, we can express the normal force: $F_N = mg - F_T \sin\theta$.

The crate is accelerating horizontally, so $\Sigma F_x = ma$.
$F_{Tx} - F_f = ma$
$F_T \cos\theta - F_f = ma$

$F_f = \mu_k F_N$

We will substitute the expression for $F_N$ into the friction equation, and then substitute the friction equation into the x-direction force equation to solve for $F_T$.

SOLUTION:

$F_f = \mu_k (mg - F_T \sin\theta)$

$F_T \cos\theta - \mu_k (mg - F_T \sin\theta) = ma$

$F_T \cos\theta - \mu_k mg + \mu_k F_T \sin\theta = ma$
$F_T \cos\theta + \mu_k F_T \sin\theta = ma + \mu_k mg$
$F_T (\cos\theta + \mu_k \sin\theta) = m(a + \mu_k g)$
$F_T = \frac{m(a + \mu_k g)}{\cos\theta + \mu_k \sin\theta}$

$F_T = \frac{12.0 \text{ kg} (1.44 \text{ m/s}^2 + 0.25 \times 9.81 \text{ m/s}^2)}{\cos(30.0^\circ) + 0.25 \times \sin(30.0^\circ)}$
$F_T = \frac{12.0 \text{ kg} (1.44 \text{ m/s}^2 + 2.4525 \text{ m/s}^2)}{0.8660 + 0.25 \times 0.5000}$
$F_T = \frac{12.0 \text{ kg} (3.8925 \text{ m/s}^2)}{0.8660 + 0.1250}$
$F_T = \frac{46.71 \text{ N}}{0.9910}$
$F_T = 47.134 \text{ N}$

$F_T = 47.1 \text{ N}$

STATEMENT:
The magnitude of the pulling force applied to the rope is $47.1 \text{ N}$.

Question 4: d) Record a 1–2-minute audio response explaining why the normal force is not equal to the weight of the crate in this situation, and describe how this affects the frictional force. Use physics terminology and refer to your FBD in your explanation.

Note: As an AI, I cannot record an audio response. However, I can provide the script for a 1-2 minute explanation that addresses all parts of your question.

(Audio Response Script)

"Hello! Let's discuss why the normal force on the crate in this problem is not equal to its weight, and how that impacts the frictional force.

Referring to our Free-Body Diagram, we have four forces acting on the crate: the force of gravity, or weight ($F_g$), pulling downwards; the normal force ($F_N$), pushing upwards from the floor; the tension force ($F_T$) from the rope, acting at an angle of 30 degrees above the horizontal; and the kinetic friction force ($F_f$), opposing the motion horizontally.

In many simpler scenarios, when an object rests on a horizontal surface and no other vertical forces are present, the normal force is indeed equal in magnitude to the force of gravity. This is because these two forces are the only ones acting vertically, and they must balance each other out to prevent vertical acceleration.

However, in this situation, the rope is pulling the crate at an angle. This means the tension force ($F_T$) has both a horizontal component ($F_T \cos\theta$) and a vertical component ($F_T \sin\theta$). Crucially, the vertical component of the tension, $F_T \sin\theta$, is directed upwards.

So, when we analyze the vertical forces, we have the upward normal force ($F_N$) and the upward vertical component of tension ($F_T \sin\theta$) counteracting the downward force of gravity ($F_g$). Since the crate is not accelerating vertically, the sum of the upward forces must equal the sum of the downward forces. Therefore, $F_N + F_T \sin\theta = F_g$.

Rearranging this equation, we find that $F_N = F_g - F_T \sin\theta$. This clearly shows that the normal force is less than the force of gravity because the upward component of the tension force is effectively 'lifting' the crate slightly off the floor, reducing the force with which the crate presses down on the surface.

Now, how does this affect the frictional force? The kinetic frictional force ($F_f$) is directly proportional to the normal force, as described by the equation $F_f = \mu_k F_N$, where $\mu_k$ is the coefficient of kinetic friction. Since the normal force ($F_N$) in this scenario is reduced (it's less than the weight of the crate), the kinetic frictional force ($F_f$) will also be reduced. A smaller normal force means less resistance to motion from friction, making it easier to accelerate the crate horizontally."