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. provide the image also

Question 1: a) Draw a fully labeled Free-Body Diagram (FBD) for the crate. Include all forces acting on the crate.

a) Free-Body Diagram (FBD)

        ^ F_N (Normal Force)
        |
        |
        |      / F_T (Tension)
        |     /
        |    / 30.0°
        |   /
        |  /
        | /
        |/
<-------O-------> F_k (Kinetic Friction)
        |
        |
        |
        v F_g (Weight)

Labels:

• F_g: Force of gravity (Weight) acting downwards.
• F_N: Normal force exerted by the floor, acting upwards, perpendicular to the surface.
• F_T: Tension force from the rope, acting at 30.0° above the horizontal.
• F_k: Kinetic friction force, acting horizontally opposite to the direction of motion.

Question 2: b) Determine the acceleration of the crate using kinematics.

b) Acceleration of the Crate

GIVEN:

• Initial velocity, v_i = 0 m/s
• Displacement, Δx = 18.0 m
• Time, Δt = 5.00 s

REQUIRED:

• Acceleration, a

ANALYSIS:
Since the crate starts from rest and moves with constant acceleration, we can use the kinematic equation:
Δx = v_i * Δt + (1/2) * a * (Δt)^2
Given v_i = 0, the equation simplifies to:
Δx = (1/2) * a * (Δt)^2
We can rearrange this to solve for a:
a = 2 * Δx / (Δt)^2

SOLUTION:
a = 2 * (18.0 m) / (5.00 s)^2
a = 36.0 m / 25.0 s^2
a = 1.44 m/s^2

STATEMENT:
The acceleration of the crate is 1.44 m/s^2.

Question 3: c) Calculate the magnitude of the pulling force applied to the rope. Show all steps and include appropriate units.

c) Magnitude of the Pulling Force

GIVEN:

• Mass, m = 12.0 kg
• Angle, θ = 30.0°
• Coefficient of kinetic friction, μ_k = 0.25
• Acceleration, a = 1.44 m/s^2 (from part b)
• Acceleration due to gravity, g = 9.81 m/s^2

REQUIRED:

• Pulling force (Tension), F_T

ANALYSIS:
We will apply Newton's Second Law in both the horizontal (x) and vertical (y) directions.
From the FBD:
Forces in x-direction: ΣF_x = F_T * cos(θ) - F_k = ma
Forces in y-direction: ΣF_y = F_N + F_T * sin(θ) - F_g = 0 (since there is no vertical acceleration)
We also know that F_g = mg and F_k = μ_k * F_N.

From the y-direction equation, solve for F_N:
F_N = F_g - F_T * sin(θ)
F_N = mg - F_T * sin(θ)

Substitute F_N into the x-direction equation:
F_T * cos(θ) - μ_k * (mg - F_T * sin(θ)) = ma
Now, solve for F_T:
F_T * cos(θ) - μ_k * mg + μ_k * F_T * sin(θ) = ma
F_T * (cos(θ) + μ_k * sin(θ)) = ma + μ_k * mg
F_T = (ma + μ_k * mg) / (cos(θ) + μ_k * sin(θ))

SOLUTION:
First, calculate mg:
F_g = (12.0 kg) * (9.81 m/s^2) = 117.72 N

Now, substitute all values into the equation for F_T:
F_T = ((12.0 kg) * (1.44 m/s^2) + (0.25) * (117.72 N)) / (cos(30.0°) + (0.25) * sin(30.0°))
F_T = (17.28 N + 29.43 N) / (0.8660 + 0.25 * 0.5000)
F_T = (46.71 N) / (0.8660 + 0.1250)
F_T = (46.71 N) / (0.9910)
F_T = 47.134 N

Rounding to two significant figures due to μ_k = 0.25:
F_T = 47 N

STATEMENT:
The magnitude of the pulling force applied to the rope is 47 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.

d) Explanation for Normal Force and Frictional Force (Audio Response Content)

"In this scenario, where a crate is pulled across a horizontal floor by a rope angled upwards, the normal force is not equal to the weight of the crate. Let's refer to the Free-Body Diagram (FBD) to understand why.

On our FBD, we have four forces acting on the crate: the downward force of gravity, or weight (F_g), the upward normal force (F_N) from the floor, the tension force (F_T) from the rope, and the kinetic friction force (F_k) opposing motion.

The key here is the tension force (F_T) from the rope, which is applied at an angle of 30.0° above the horizontal. This means the tension force has two components: a horizontal component (F_T * cos(θ)) that pulls the crate forward, and a vertical component (F_T * sin(θ)) that pulls the crate upwards.

Because there is an upward vertical component of the tension force, this component effectively helps to support the crate against gravity. The floor, therefore, does not need to exert as much upward normal force to counteract the entire weight of the crate. In the vertical direction, the sum of forces is zero because the crate is not accelerating vertically. So, the upward normal force (F_N) plus the upward vertical component of tension (F_T * sin(θ)) must balance the downward force of gravity (F_g). Mathematically, this is expressed as F_N + F_T * sin(θ) - F_g = 0, which means F_N = F_g - F_T * sin(θ). Since F_T * sin(θ) is a positive value, the normal force (F_N) will be less than the weight (F_g).

Now, let's consider how this affects the frictional force. The kinetic frictional force (F_k) is directly proportional to the normal force, given by the equation F_k = μ_k * F_N, where μ_k is the coefficient of kinetic friction. Since we've established that the normal force (F_N) is reduced because of the upward component of the pulling force, the kinetic frictional force (F_k) will also be reduced. A smaller normal force results in a smaller frictional force opposing the motion. This reduction in friction makes it easier to accelerate the crate or requires less horizontal effort to move it compared to pulling it horizontally or pushing it downwards."