To calculate the work done by the movers in pushing the crate, we can use the formula for work done against friction:

$\text{Work} = \text{Friction Force} \times \text{Distance}$

First, we need to determine the frictional force. The frictional force ($F_f$) can be calculated using the equation:

$F_f = \mu \times F_n$

where:

• $\mu$ is the coefficient of friction (0.400),
• $F_n$ is the normal force.

For a crate on a horizontal surface, the normal force equals the weight of the crate:

$F_n = m \times g$

Here:

• $m = 46.0 \, \text{kg}$ (mass of the crate),
• $g = 9.81 \, \text{m/s}^2$ (acceleration due to gravity).

Calculating the normal force:

$F_n = 46.0 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 450.26 \, \text{N}$

Now, calculating the frictional force:

$F_f = 0.400 \times 450.26 \, \text{N} = 180.10 \, \text{N}$

Now, we can calculate the work done over the distance of 15.3 m:

$\text{Work} = 180.10 \, \text{N} \times 15.3 \, \text{m} = 2754.53 \, \text{J}$

Thus, the work done by the movers is approximately 2755 J.

Would you like more details or do you have any questions? Here are some related questions you might consider:

1. How would the work change if the coefficient of friction increased to 0.500?
2. What is the effect of increasing the mass of the crate on the work done?
3. How can the work done be calculated if the crate was accelerating instead?
4. What other factors could affect the frictional force in this scenario?
5. How would the work done change if the distance were doubled?

Tip: When dealing with friction, always ensure you correctly identify the normal force, as it directly impacts the frictional force.