To determine the work done by Don in pushing the barrel up the ramp, we use the formula for work:

$\text{Work} = \text{Force} \times \text{Distance} \times \cos(\theta)$

where:

• $\text{Force} = 534 \, \text{N}$
• $\text{Distance} = 1.5 \, \text{m}$
• $\theta = 25^\circ$ (angle of the ramp from the horizontal)

Given that the force is acting parallel to the ramp surface, the angle $\theta$ in this context is not needed for cosine adjustment because the force component along the ramp is already given as 534 N. Thus, we can simplify the work calculation as:

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

Substitute the given values:

$\text{Work} = 534 \, \text{N} \times 1.5 \, \text{m}$

$\text{Work} = 801 \, \text{J}$

So, the work done by Don is:

$801$

Would you like more details or have any questions?

Here are 8 related questions for further practice:

1. What is the work done if the force was 400 N instead of 534 N?
2. How does the angle of the ramp affect the work done if the force were not parallel to the ramp?
3. If the ramp length were doubled, how much work would be done for the same force?
4. How much work is done if the ramp angle is increased to 30° with the same force and ramp length?
5. What is the definition of work in the context of physics?
6. How does friction affect the work done when pushing an object up a ramp?
7. Calculate the work done if the distance is halved and the force remains the same.
8. If Don pushed the barrel down the ramp with the same force, how would the work done compare?

Tip: Always ensure the force component you use in the work calculation is in the direction of the displacement.