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

To solve the problem, we first calculate the work done in joules and then convert it into millielectronvolts (meV).

1. Calculate Work Done in Joules:

Work done is given by: \[ W = F \times d \times \cos \theta \] Where:
\(F = 180 \, \text{N}\), \(d = 25 \, \text{m}\), and \(\theta = 45^\circ\).
Using \(\cos 45^\circ = \frac{\sqrt{2}}{2} \approx 0.707\): \[ W = 180 \times 25 \times 0.707 = 3181.5 \, \text{J} \]

2. Convert Joules to Millielectronvolts (meV):

- 1 electronvolt (eV) = \(1.6 \times 10^{-19}\) joules (J)
- 1 millielectronvolt (meV) = \(10^{-3}\) eV

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

Final Answer:

The work done to slide the crate is approximately 1.988 × 10²⁵ millielectronvolts (meV).