2. A boy pulls a 26.0-kg box with a 170-N force at 38° above a horizontal surface. If the coefficient of kinetic friction between the box and the what horizontal surface is 0.18 and the box is pulled a distance of 28.0m cci16 à is the net work done on the box?
2. A boy pulls a 26.0-kg box with a 170-N force at 38° above a horizontal surface. If the coefficient of kinetic friction between the box and the what horizontal surface is 0.18 and the box is pulled a distance of 28.0m cci16 à is the net work done on the box?
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![**Problem 2: Calculation of Net Work Done on a Box**
A boy pulls a 26.0-kg box with a 170-N force at 38° above a horizontal surface. If the coefficient of kinetic friction between the box and the horizontal surface is 0.18 and the box is pulled a distance of 28.0 m, what is the net work done on the box?
**Solution:**
To find the net work done on the box, follow the steps below:
1. **Determine the horizontal and vertical components of the pulling force:**
- Horizontal component of the force (\( F_x \)):
\[
F_x = F \cos(\theta) = 170 \, \text{N} \cdot \cos(38^\circ)
\]
- Vertical component of the force (\( F_y \)):
\[
F_y = F \sin(\theta) = 170 \, \text{N} \cdot \sin(38^\circ)
\]
2. **Calculate the normal force (\( F_N \)):**
The normal force is the force exerted by the surface to support the weight of the box, adjusted for the vertical component of the applied force.
\[
F_N = mg - F_y
\]
where \( m \) is the mass of the box (26.0 kg) and \( g \) is the acceleration due to gravity (9.8 m/s²).
3. **Calculate the frictional force (\( F_f \)):**
The frictional force depends on the normal force and the coefficient of kinetic friction (\( \mu_k \)).
\[
F_f = \mu_k F_N
\]
4. **Determine the net force acting on the box in the horizontal direction:**
The net force is the horizontal component of the pulling force minus the frictional force.
\[
F_{\text{net}} = F_x - F_f
\]
5. **Calculate the work done by the net force:**
The work done by the net force is the product of the net force and the distance \( d \) over which the force is applied.
\[
W_{\text{net}} = F_{\text{net}} \cdot d
\]
where \( d \) is 28.0 m](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcd0a9c9c-b0c8-47bb-88dc-3ac08f167320%2Febb36f91-0ff2-4bff-95de-0d97d473d0ba%2F8poz28r_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem 2: Calculation of Net Work Done on a Box**
A boy pulls a 26.0-kg box with a 170-N force at 38° above a horizontal surface. If the coefficient of kinetic friction between the box and the horizontal surface is 0.18 and the box is pulled a distance of 28.0 m, what is the net work done on the box?
**Solution:**
To find the net work done on the box, follow the steps below:
1. **Determine the horizontal and vertical components of the pulling force:**
- Horizontal component of the force (\( F_x \)):
\[
F_x = F \cos(\theta) = 170 \, \text{N} \cdot \cos(38^\circ)
\]
- Vertical component of the force (\( F_y \)):
\[
F_y = F \sin(\theta) = 170 \, \text{N} \cdot \sin(38^\circ)
\]
2. **Calculate the normal force (\( F_N \)):**
The normal force is the force exerted by the surface to support the weight of the box, adjusted for the vertical component of the applied force.
\[
F_N = mg - F_y
\]
where \( m \) is the mass of the box (26.0 kg) and \( g \) is the acceleration due to gravity (9.8 m/s²).
3. **Calculate the frictional force (\( F_f \)):**
The frictional force depends on the normal force and the coefficient of kinetic friction (\( \mu_k \)).
\[
F_f = \mu_k F_N
\]
4. **Determine the net force acting on the box in the horizontal direction:**
The net force is the horizontal component of the pulling force minus the frictional force.
\[
F_{\text{net}} = F_x - F_f
\]
5. **Calculate the work done by the net force:**
The work done by the net force is the product of the net force and the distance \( d \) over which the force is applied.
\[
W_{\text{net}} = F_{\text{net}} \cdot d
\]
where \( d \) is 28.0 m
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