As shown in the figure below, a box of mass

*m* = 54.0 kg

(initially at rest) is pushed a distance

*d* = 67.0 m

across a rough warehouse floor by an applied force of

*FA* = 206 N

directed at an angle of 30.0° below the horizontal. The coefficient of kinetic friction between the floor and the box is 0.100. Determine the following. (For parts (a) through (d), give your answer to the nearest multiple of 10.)Find the work done by the force of friction. 
Calculate the net work on the box by finding the sum of all the works done by each individual force.Now find the net work by first finding the net force on the box, then finding the work done by this net force.

### Problem Breakdown:

We are given the following parameters:
- Mass of the box: \( m = 54.0 \, \text{kg} \)
- Distance pushed: \( d = 67.0 \, \text{m} \)
- Applied force: \( F_A = 206 \, \text{N} \) at an angle of 30.0° below the horizontal.
- Coefficient of kinetic friction: \( \mu_k = 0.100 \)

### (a) Work Done by the Force of Friction

First, we need to find the force of friction \( F_f \) and the work done by it.

1. **Normal Force**:
   The normal force \( F_N \) is affected by both the weight of the box and the vertical component of the applied force.

   The weight of the box:
   \[
   F_g = mg = 54.0 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 529.2 \, \text{N}
   \]

   The vertical component of the applied force:
   \[
   F_{A_y} = F_A \sin(30.0^\circ) = 206 \times 0.5 = 103 \, \text{N}
   \]

   The total normal force is then:
   \[
   F_N = F_g - F_{A_y} = 529.2 - 103 = 426.2 \, \text{N}
   \]

2. **Friction Force**:
   The frictional force is given by:
   \[
   F_f = \mu_k F_N = 0.100 \times 426.2 = 42.62 \, \text{N}
   \]

3. **Work Done by Friction**:
   The work done by the force of friction is:
   \[
   W_f = -F_f \times d = -42.62 \times 67.0 = -2856.54 \, \text{J}
   \]

   Rounding to the nearest multiple of 10:
   \[
   W_f = -2860 \, \text{J}
   \]

### (b) Net Work by Summing Individual Works

The forces acting on the box include the applied force, the gravitational force, the normal force, and the friction force.

1. **Work Done by Applied Force**:
   Only the horizontal component of the applied force does work:
   \[
   F_{A_x} = F_A \cos(30.0^\circ) = 206 \times 0.866 = 178.1 \, \text{N}
   \]
   The work done by this horizontal component is:
   \[
   W_A = F_{A_x} \times d = 178.1 \times 67.0 = 11932.7 \, \text{J}
   \]
   Rounding to the nearest multiple of 10:
   \[
   W_A = 11930 \, \text{J}
   \]

2. **Work Done by Gravitational Force**:
   The work done by gravity is 0 because the box is moving horizontally.

3. **Work Done by Normal Force**:
   The normal force also does no work because it acts perpendicular to the motion.

4. **Net Work**:
   The net work by summing the individual works is:
   \[
   W_{\text{net}} = W_A + W_f = 11930 + (-2860) = 9070 \, \text{J}
   \]

### (c) Net Work by Finding Net Force First

1. **Net Horizontal Force**:
   The net horizontal force is the horizontal component of the applied force minus the frictional force:
   \[
   F_{\text{net}} = F_{A_x} - F_f = 178.1 - 42.62 = 135.48 \, \text{N}
   \]

2. **Net Work**:
   The net work done by this force is:
   \[
   W_{\text{net}} = F_{\text{net}} \times d = 135.48 \times 67.0 = 9087.16 \, \text{J}
   \]
   Rounding to the nearest multiple of 10:
   \[
   W_{\text{net}} = 9090 \, \text{J}
   \]

### Summary of Results:
- (a) Work done by friction: \( W_f = -2860 \, \text{J} \)
- (b) Net work by summing individual works: \( W_{\text{net}} = 9070 \, \text{J} \)
- (c) Net work by finding net force: \( W_{\text{net}} = 9090 \, \text{J} \)

Would you like further clarifications or details on any part of the solution?

---

**Relative Questions:**
1. How does the angle of the applied force affect the normal force and friction?
2. What would happen to the work done by friction if the coefficient of friction increased?
3. How would the results change if the applied force was horizontal?
4. What is the significance of the work done by forces that act perpendicular to the motion?
5. How does the net work relate to the box's change in kinetic energy?

**Tip:** When calculating work, only the force component in the direction of motion contributes to the work done.

As shown in the figure below, a box of mass *m* = 54.0 kg (initially at rest) is pushed a distance *d* = 67.0 m across a rough warehouse floor by an applied force of *FA* = 206 N directed at an angle of 30.0° below the horizontal. The coefficient of kinetic friction between the floor and the box is 0.100. Determine the following. (For parts (a) through (d), give your answer to the nearest multiple of 10.) Find the work done by the force of friction. Calculate the net work on the box by finding the sum of all the works done by each individual force. Now find the net work by first finding the net force on the box, then finding the work done by this net force.

Learn how to calculate the work done by friction and the net work on a 54 kg box pushed across a rough surface. This problem covers key physics concepts such as the work-energy principle, forces in motion, and kinetic friction. Suitable for Grades 11-12.

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