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 \(45^\circ\) from the horizontal?

• A. \( 3.18198 \times 10^3 \, {J} \)
• B. \( 3.18198 \times 10^2 \, {J} \)
• C. \( 3.4341 \times 10^3 \, {J} \)
• D. \( 3.4341 \times 10^4 \, {J} \)

Hint:

To calculate work, remember the formula \( {Work} = F \cdot d \cdot \cos(\theta) \), where \( F \) is the force, \( d \) is the displacement, and \( \theta \) is the angle between the direction of force and displacement.

Answer 1

The Correct Option is A

The work done is given by the formula: \[ {Work} = F \cdot d \cdot \cos(\theta) \] where:
\( F = 180 \, {N} \) is the force applied,
\( d = 25 \, {m} \) is the distance moved,
\( \theta = 45^\circ \) is the angle between the force and the horizontal direction.
Substituting the values into the equation: \[ {Work} = 180 \cdot 25 \cdot \cos(45^\circ) \] Since \( \cos(45^\circ) = \frac{\sqrt{2}}{2} \), we get: \[ {Work} = 180 \cdot 25 \cdot \frac{\sqrt{2}}{2} = 3181.98 \, {J} \]