4. A crate is pulled up a ramp a distance of 20 m with a force of F = [800, 500]. The crate ends up 2 m higher above the ground than where it started.

1. Break Down the Problem

To analyze the situation, we need to:

1. Determine the work done on the crate by the force F.
2. Calculate the gravitational potential energy change when the crate is raised.
3. Verify any relations between the work done and the change in potential energy.

2. Relevant Concepts

1. Work Done by a Force: The work $W$ done by a force is given by the dot product of the force vector and the displacement vector: $W = \mathbf{F} \cdot \mathbf{d}$ where $\mathbf{F} = [800, 500]$ and $\mathbf{d} = [20, 2]$.

2. Gravitational Potential Energy: The change in potential energy $\Delta PE$ is calculated as: $\Delta PE = mgh$ where $m$ is the mass of the crate (not given), $g$ is the acceleration due to gravity (approximately $9.81 \, \text{m/s}^2$), and $h$ is the height change.

3. Analysis and Detail

1. Calculating the Work Done: $W = \mathbf{F} \cdot \mathbf{d} = (800, 500) \cdot (20, 2) = 800 \times 20 + 500 \times 2 = 16000 + 1000 = 17000 \, \text{J}$

2. Calculating the Change in Potential Energy: To calculate $\Delta PE$, we need the mass of the crate. However, we can express it generally. $h = 2 \, \text{m}, \quad g = 9.81 \, \text{m/s}^2$ $\Delta PE = mgh = m \cdot 9.81 \cdot 2 = 19.62m \, \text{J}$

4. Verify and Summarize

The work done $W$ is $17000 \, \text{J}$, while the change in potential energy $\Delta PE = 19.62m \, \text{J}$.

If we equate the work done to the change in potential energy to find the mass: $17000 = 19.62m$ $m = \frac{17000}{19.62} \approx 867.2 \, \text{kg}$

Final Answer

The work done on the crate is $17000 \, \text{J}$, and the approximate mass of the crate is $867.2 \, \text{kg}$.