Let's solve each of the problems one by one using the formula for work:

$W = F \cdot d$

Where:

• $W$ is the work in joules (J)
• $F$ is the force in newtons (N)
• $d$ is the displacement in meters (m)

1. $F = 90 \, \text{N}, d = 5 \, \text{m}, W = ?$

Using the formula:

$W = 90 \, \text{N} \times 5 \, \text{m} = 450 \, \text{J}$

So, $W = 450 \, \text{J}$.

---

2. $W = ?, F = 62.6 \, \text{N}, d = 13 \, \text{m}$

Using the formula again:

$W = 62.6 \, \text{N} \times 13 \, \text{m} = 813.8 \, \text{J}$

So, $W = 813.8 \, \text{J}$.

---

3. $F = 6 \, \text{N}, W = 72 \, \text{J}, d = ?$

Rearrange the formula to solve for $d$:

$d = \frac{W}{F} = \frac{72 \, \text{J}}{6 \, \text{N}} = 12 \, \text{m}$

So, $d = 12 \, \text{m}$.

---

4. $W = 13.2 \, \text{J}, F = 2 \, \text{N}, d = ?$

Rearrange the formula to solve for $d$:

$d = \frac{W}{F} = \frac{13.2 \, \text{J}}{2 \, \text{N}} = 6.6 \, \text{m}$

So, $d = 6.6 \, \text{m}$.

---

5. $W = 120 \, \text{J}, d = 24 \, \text{m}, F = ?$

Rearrange the formula to solve for $F$:

$F = \frac{W}{d} = \frac{120 \, \text{J}}{24 \, \text{m}} = 5 \, \text{N}$

So, $F = 5 \, \text{N}$.

---

6. $W = 136 \, \text{J}, d = 27.2 \, \text{m}, F = ?$

Rearrange the formula to solve for $F$:

$F = \frac{W}{d} = \frac{136 \, \text{J}}{27.2 \, \text{m}} = 5 \, \text{N}$

So, $F = 5 \, \text{N}$.

---

Would you like further details or have any questions on any of the solutions?

Further Questions:

1. How would the work change if the force was doubled?
2. How does the distance traveled affect the work done when the force is constant?
3. What is the relationship between work and energy in physics?
4. What units are typically used for force, distance, and work in SI?
5. Can work be negative? If so, when?

Tip: When calculating work, ensure you use consistent units (newtons for force, meters for distance, joules for work) to avoid conversion errors.