Question

How much work is done to slide a crate for a distance of 25 m along a loading dock by pulling on it with a 180 N force where the dock is at an angle of $45^\circ$45∘ from the horizontal?

• A. 1.188 × 10²⁵ MeV
• B. 1.288 × 10²⁵ MeV
• C. 1.588 × 10²⁵ MeV
• D. 1.988 × 10²⁵ MeV

Answer 1

The Correct Option is D

The solution involves two steps: first, calculating the work done in joules, and second, converting this value to millielectronvolts (meV).

1. Work Done Calculation (Joules):

The formula for work done is:\[W = F \times d \times \cos \theta\]Given values are:\(F = 180 \, \text{N}\), \(d = 25 \, \text{m}\), and \(\theta = 45^\circ\).Since \(\cos 45^\circ \approx 0.707\):\[W = 180 \times 25 \times 0.707 = 3181.5 \, \text{J}\]

2. Conversion to Millielectronvolts (meV):

Conversion factors:- 1 eV = \(1.6 \times 10^{-19}\) J
- 1 meV = \(10^{-3}\) eV

First, convert joules to eV:\[W_{\text{eV}} = \frac{3181.5}{1.6 \times 10^{-19}} \approx 1.988 \times 10^{22} \, \text{eV}\]Next, convert eV to meV:\[W_{\text{meV}} = 1.988 \times 10^{22} \times 10^{3} = 1.988 \times 10^{25} \, \text{meV}\]

Final Result:

The calculated work done in sliding the crate is approximately 1.988 × 10²⁵ meV.