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To solve this problem, we will approach each part methodically:

### 1. **Work done by each force on the block**
There are two main forces doing work on the block:
   - **Tension in the rope** pulling the block.
   - **Frictional force** opposing the block's motion.

#### a. Work done by the **tension force**:
The tension force \( T \) is applied at an angle of \( 28^\circ \) above the horizontal. To find the horizontal component of the tension force, we use:

\[
T_x = T \cdot \cos(\theta)
\]
Where \( T = 1000 \, \text{N} \) and \( \theta = 28^\circ \).

The work done by the tension force is given by:

\[
W_T = T_x \cdot d = T \cdot \cos(\theta) \cdot d
\]
Where \( d = 10 \, \text{m} \) is the displacement.

Now let's compute the horizontal component of the force and the work done:
\[
T_x = 1000 \cdot \cos(28^\circ)
\]
Using \( \cos(28^\circ) \approx 0.8829 \):

\[
T_x = 1000 \cdot 0.8829 = 882.9 \, \text{N}
\]
Now, the work done by the tension force:
\[
W_T = 882.9 \, \text{N} \cdot 10 \, \text{m} = 8829 \, \text{J}
\]

#### b. Work done by **frictional force**:
The frictional force opposes the motion and is given by:

\[
f_k = \mu_k \cdot N
\]
Where:
- \( \mu_k = 0.275 \) is the coefficient of kinetic friction,
- \( N \) is the normal force, and \( N = M \cdot g \),
- \( M = 225 \, \text{kg} \), and \( g = 9.81 \, \text{m/s}^2 \).

So, the frictional force is:

\[
N = 225 \cdot 9.81 = 2207.25 \, \text{N}
\]
\[
f_k = 0.275 \cdot 2207.25 = 606.49 \, \text{N}
\]

The work done by friction (which is in the opposite direction of motion) is:

\[
W_f = -f_k \cdot d = -606.49 \, \text{N} \cdot 10 \, \text{m} = -6064.9 \, \text{J}
\]

### 2. **Total work done on the block**
The total work done on the block is the sum of the work done by the tension and the work done by the frictional force:

\[
W_{\text{total}} = W_T + W_f = 8829 \, \text{J} - 6064.9 \, \text{J} = 2764.1 \, \text{J}
\]

### Final answers:
1. **Work done by tension force**: \( W_T = 8829 \, \text{J} \)
2. **Work done by friction force**: \( W_f = -6064.9 \, \text{J} \)
3. **Total work done on the block**: \( W_{\text{total}} = 2764.1 \, \text{J} \)

---

Would you like more details on any part of this solution? Here are some related questions you might explore:

1. How would the work done change if the angle of the rope was increased to \( 45^\circ \)?
2. What happens to the total work if the coefficient of kinetic friction is reduced by half?
3. How does the normal force affect the frictional work done in this problem?
4. How much work would the tension force do if there were no friction?
5. What if the block were moved over a distance of 20 m instead of 10 m?

**Tip**: Always break the forces into components (horizontal and vertical) when dealing with forces acting at an angle.

A rope pulls on a massive block of mass M = 225 kg across a rough floor with a coefficient of kinetic friction µk = 0.275. The rope exerts a force of 1000 N at 28° above the horizontal. When the block has moved 10 m from its initial position, what is the work done by each force on the block? What is the total work done on the block?

Learn how to calculate the work done by a rope pulling a block across a rough floor with friction. Includes step-by-step solutions for calculating the work done by tension and friction forces using work-energy principles.

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