A 55-kg box is being pushed a distance of 7.0 m across the floor by a force `vec(p)` whose magnitude is 150 N. The force `vec(P)` is parallel to the displacement of the box. The coefficient of kinetic friction is 0.25. Determine the work done on the box by each of the four forces that act on the box. Be sure to include the proper plus or minus sign for the work done by each force. ( The four forces and the force P, the frictional force f, the normal force N, and the force due to gravity mg. )

Options

A. `{:(""W_(p),""W_(f),""W_(N),""W_(mg)),(1.0xx10^(3)J,-940J,-3.8xx10^(3)J,3.8xx10^(3)J):}`
B. `{:(""W_(p),""W_(f),""W_(N),""W_(mg)),(1.0xx10^(3),-940J,"zero"J,"zeroJ"):}`
C. `{:(""W_(p),""W_(f),""W_(N),""W_(mg)),("zeroJ",940J,-3.8xx10^(3)J,3.8xx10^(3)J):}`
D. `{:(""W_(p),""W_(f),""W_(N),""W_(mg)),(-1.0xx10^(3)J,"zeroJ","zeroJ",-3.8xx10^(3)J):}`

Answer

To solve the problem, we need to calculate the work done by each of the four forces acting on the box: the applied force \( \vec{P} \), the frictional force \( \vec{F} \), the normal force \( \vec{N} \), and the gravitational force \( \vec{W} \).

**Step 1: Calculate the work done by the applied force \( \vec{P} \)**

The work done by a force is given by the formula:
\[ W = F \cdot d \cdot \cos(\theta) \]
where:
- \( F \) is the magnitude of the force,
- \( d \) is the displacement,
- \( \theta \) is the angle between the force and the direction of displacement.

In this case, the applied force \( \vec{P} \) has a magnitude of 150 N and is parallel to the displacement (7.0 m), so \( \theta = 0^\circ \).

\[ W_P = 150 \, \text{N} \cdot 7.0 \, \text{m} \cdot \cos(0^\circ) = 150 \cdot 7 \cdot 1 = 1050 \, \text{J} \]

**Step 2: Calculate the frictional force \( \vec{F} \)**

The frictional force can be calculated using the formula:
\[ F = \mu \cdot N \]
where \( \mu \) is the coefficient of kinetic friction and \( N \) is the normal force. The normal force \( N \) is equal to the weight of the box, which is given by:
\[ N = mg \]
where \( m = 55 \, \text{kg} \) and \( g = 9.8 \, \text{m/s}^2 \).

Calculating \( N \):
\[ N = 55 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 539 \, \text{N} \]

Now, calculating the frictional force:
\[ F = 0.25 \cdot 539 \, \text{N} = 134.75 \, \text{N} \]

**Step 3: Calculate the work done by the frictional force \( \vec{F} \)**

The frictional force acts opposite to the direction of displacement, so \( \theta = 180^\circ \).

\[ W_F = F \cdot d \cdot \cos(180^\circ) = 134.75 \, \text{N} \cdot 7.0 \, \text{m} \cdot (-1) = -943.25 \, \text{J} \]

**Step 4: Calculate the work done by the normal force \( \vec{N} \)**

The normal force acts perpendicular to the displacement, so \( \theta = 90^\circ \).

\[ W_N = N \cdot d \cdot \cos(90^\circ) = 539 \, \text{N} \cdot 7.0 \, \text{m} \cdot 0 = 0 \, \text{J} \]

**Step 5: Calculate the work done by the gravitational force \( \vec{W} \)**

The gravitational force also acts perpendicular to the displacement, so \( \theta = 90^\circ \).

\[ W_W = W \cdot d \cdot \cos(90^\circ) = mg \cdot d \cdot 0 = 0 \, \text{J} \]

**Final Results:**

- Work done by the applied force \( W_P = 1050 \, \text{J} \)
- Work done by the frictional force \( W_F = -943.25 \, \text{J} \)
- Work done by the normal force \( W_N = 0 \, \text{J} \)
- Work done by the gravitational force \( W_W = 0 \, \text{J} \)

### Summary of Work Done by Each Force:
- \( W_P = +1050 \, \text{J} \)
- \( W_F = -943.25 \, \text{J} \)
- \( W_N = 0 \, \text{J} \)
- \( W_W = 0 \, \text{J} \)