To determine the work done by John, we use the formula for work:

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

where:

• $W$ is the work done,
• $F$ is the force applied,
• $d$ is the distance moved along the direction of the force,
• $\theta$ is the angle between the force and the direction of movement.

In this problem:

• $F = 80$ newtons,
• $d = 10$ meters,
• $\theta = 0^\circ$ because the force applied by John is parallel to the direction of the movement up the ramp (since the force is pushing up the ramp and the box moves up the ramp).

Since the angle $\theta$ between the force and the direction of movement is $0^\circ$, $\cos(0^\circ) = 1$.

Thus, the work done $W$ is:

$W = 80 \, \text{N} \cdot 10 \, \text{m} \cdot \cos(0^\circ)$

$W = 80 \, \text{N} \cdot 10 \, \text{m} \cdot 1$

$W = 800 \, \text{Joules}$

Rounded to the nearest whole number, the amount of work John does is:

$800$

Would you like any details or have any questions?

Here are 8 related questions to expand your understanding:

1. How would the work done change if the angle of the ramp increased to 30° but the force and distance remained the same?
2. What is the significance of the cosine component in the work formula?
3. If friction were present, how would that affect the calculation of work done?
4. How does the work-energy principle relate to this problem?
5. What if the force applied was not constant; how would you approach the problem then?
6. How can we determine the kinetic energy of the box after moving 10 meters?
7. What role does gravity play in this scenario?
8. How would the work done differ if John pulled the box up the ramp instead of pushing it?

Tip: When calculating work done, always ensure that the direction of force and movement are correctly aligned to use the cosine component accurately.