Concept: Work done by a constant force is a scalar quantity that measures the energy transferred to or from an object when it is moved by that force. It depends on the magnitude of the force, the magnitude of the displacement, and the angle between the force and displacement vectors.
Step 1: Recall the Formula for Work Done
The work done (\(W\)) by a constant force (\(F\)) that causes a displacement (\(d\)) is given by:
\[ W = F d \cos \theta \]
where:
\(F\) is the magnitude of the constant force applied.
\(d\) is the magnitude of the displacement of the object.
\(\theta\) is the angle between the direction of the force vector \(\vec{F}\) and the direction of the displacement vector \(\vec{d}\).
Step 2: Identify the Condition for Maximum Work
For the work done (\(W\)) to be maximum, given that \(F\) and \(d\) are constant magnitudes, the value of \(\cos \theta\) must be maximum.
Step 3: Determine the Maximum Value of \(\cos \theta\)
The cosine function, \(\cos \theta\), ranges from -1 to +1.
The maximum value of \(\cos \theta\) is \(1\).
Step 4: Find the Angle \(\theta\) for which \(\cos \theta\) is Maximum
The cosine function \(\cos \theta = 1\) when the angle \(\theta = 0^\circ\).
Step 5: Calculate Maximum Work and Evaluate Options
If \(\cos \theta = 1\) (when \(\theta = 0^\circ\)), the work done is:
\[ W_{\text{max}} = F d (1) = F d \]
Let's check the \(\cos \theta\) values for the given options:
For \(\theta = 60^\circ\): \(\cos 60^\circ = 0.5\). So, \(W = 0.5 Fd\).
For \(\theta = 45^\circ\): \(\cos 45^\circ \approx 0.707\). So, \(W \approx 0.707 Fd\).
For \(\theta = 30^\circ\): \(\cos 30^\circ \approx 0.866\). So, \(W \approx 0.866 Fd\).
For \(\theta = 0^\circ\): \(\cos 0^\circ = 1\). So, \(W = Fd\).
Comparing these values, \(Fd\) (when \(\theta = 0^\circ\)) is the largest.
Therefore, maximum work is done when the angle between force and displacement is \(0^\circ\).