To solve for the total work done on the crate, we use the formula:

$W_{\text{total}} = W_1 + W_f$

where:

• $W_1$ is the work done by the applied force,
• $W_f$ is the work done by the frictional force.

Step 1: Work done by the applied force

The work done by a force is given by:

$W = F \cdot d \cdot \cos(\theta)$

where:

• $F$ is the magnitude of the applied force (88.4 N),
• $d$ is the distance over which the force acts (11.9 m),
• $\theta$ is the angle between the force and the direction of motion (65.5°).

Substitute the values:

$W_1 = 88.4 \, \text{N} \cdot 11.9 \, \text{m} \cdot \cos(65.5^\circ)$

Using a calculator:

$\cos(65.5^\circ) \approx 0.4226$

Now calculate $W_1$:

$W_1 = 88.4 \cdot 11.9 \cdot 0.4226 \approx 440.32 \, \text{J}$

Step 2: Work done by the frictional force

The work done by the frictional force is also calculated using the work formula:

$W_f = F_f \cdot d \cdot \cos(180^\circ)$

Since the frictional force opposes the direction of motion, the angle is 180°, and $\cos(180^\circ) = -1$. Thus:

$W_f = 23.2 \, \text{N} \cdot 11.9 \, \text{m} \cdot (-1)$

$W_f = -23.2 \cdot 11.9 = -275.28 \, \text{J}$

Step 3: Total work done

Now, calculate the total work:

$W_{\text{total}} = W_1 + W_f = 440.32 \, \text{J} + (-275.28 \, \text{J})$

$W_{\text{total}} = 440.32 \, \text{J} - 275.28 \, \text{J} = 165.04 \, \text{J}$

Final Answer:

The total amount of work done is 165.04 joules.

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Would you like more details on any of the steps, or do you have any other questions about this problem?

Here are some related questions:

1. How does the angle of the applied force affect the work done?
2. What would happen if there were no frictional force?
3. How is work related to energy in this scenario?
4. What if the applied force was at a different angle, say 45 degrees?
5. How do you calculate the net force acting on the crate in this case?

Tip: When calculating work, always consider the direction of the applied force and the friction to determine whether the work is positive or negative!