https://file.aiquickdraw.com/mathbot/file/1729266576561tvrqklzj.jpg

Let's break this problem down step by step.

### Given:
- Mass of the crate \( m = 26.5 \, \text{kg} \)
- Force applied \( F = 230.25 \, \text{N} \)
- Angle of the force above the horizontal \( \theta = 17^\circ \)
- Coefficient of kinetic friction \( \mu_k = 0.25 \)
- The time interval \( t = 2.3 \, \text{seconds} \)
- The crate starts from rest, so initial velocity \( v_0 = 0 \, \text{m/s} \)

We will now calculate each of the required values:

### 1. Normal Force:
The normal force \( N \) is affected by both the gravitational force and the vertical component of the applied force. The equation for the normal force is:

\[
N = mg - F \sin(\theta)
\]

Where:
- \( g = 9.81 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( \sin(\theta) \) is the vertical component of the applied force.

Substituting the known values:

\[
N = (26.5)(9.81) - 230.25 \sin(17^\circ)
\]
\[
N \approx 259.965 - 67.345
\]
\[
N \approx 192.62 \, \text{N}
\]

### 2. Kinetic Friction Force:
The kinetic friction force \( F_k \) is given by:

\[
F_k = \mu_k N
\]

Substituting the known values:

\[
F_k = 0.25 \times 192.62
\]
\[
F_k \approx 48.16 \, \text{N}
\]

### 3. Acceleration:
The net force acting on the crate in the horizontal direction is the horizontal component of the applied force minus the kinetic friction force. The net force is:

\[
F_{\text{net}} = F \cos(\theta) - F_k
\]

The acceleration is then given by Newton's second law:

\[
a = \frac{F_{\text{net}}}{m}
\]

Substituting the values:

\[
F_{\text{net}} = 230.25 \cos(17^\circ) - 48.16
\]
\[
F_{\text{net}} \approx 220.47 - 48.16
\]
\[
F_{\text{net}} \approx 172.31 \, \text{N}
\]

Now calculate the acceleration:

\[
a = \frac{172.31}{26.5}
\]
\[
a \approx 6.50 \, \text{m/s}^2
\]

### 4. Final Velocity:
The final velocity can be found using the kinematic equation:

\[
v = v_0 + at
\]

Since the initial velocity \( v_0 = 0 \), we have:

\[
v = (6.50)(2.3)
\]
\[
v \approx 14.95 \, \text{m/s}
\]

### 5. Displacement during this time:
The displacement can be calculated using the kinematic equation:

\[
d = v_0 t + \frac{1}{2} a t^2
\]

Again, with \( v_0 = 0 \), this simplifies to:

\[
d = \frac{1}{2} (6.50)(2.3^2)
\]
\[
d \approx \frac{1}{2} \times 6.50 \times 5.29
\]
\[
d \approx 17.19 \, \text{m}
\]

### Final Answers:
- Normal force: \( 192.62 \)
- Kinetic friction force: \( 48.16 \)
- Acceleration: \( 6.50 \)
- Final velocity: \( 14.95 \)
- Displacement: \( 17.19 \)

Let me know if you need further clarification or additional details on any of these calculations! Here are some related questions to consider:

1. How does increasing the angle affect the normal force?
2. If the coefficient of friction was higher, how would it change the acceleration?
3. What would happen to the displacement if the mass of the crate was doubled?
4. How does the applied force component parallel to the floor change with different angles?
5. How would the solution change if there was no friction at all?

**Tip:** Always break down forces into their components to simplify physics problems involving angles!

A 26.5-kg crate is resting on a horizontal floor. A worker applies a force of 230.25 N at an angle of 17 degrees above the horizontal to move the crate. The coefficient of kinetic friction between the crate and the floor is 0.25. The crate starts from rest and moves for 2.3 seconds due to the pulling force. Determine the normal force, kinetic friction force, acceleration, final velocity, and displacement during this time.

Solve a physics problem involving a 26.5-kg crate on a horizontal floor. A force of 230.25 N is applied at a 17-degree angle above the horizontal, and the crate experiences kinetic friction with a coefficient of 0.25. Learn how to calculate the normal force, kinetic friction, acceleration, final velocity, and displacement over a time of 2.3 seconds.

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