CLASSICAL MECHANICS & THERMODYNAMICS - Chapter 3, Exercise 1 Solution
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Work Done at an Angle (W = Fd cos(theta))

PROBLEM
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A mover pushes a sofa 6 m across a room with a constant horizontal
force of 80 N. Calculate the work done. If instead the mover had
pushed at a 30 degree angle above horizontal with the same 80 N
force, would the work done over the same 6 m horizontal displacement
be more, less, or the same? Explain using the W = Fd cos(theta)
formula.

SOLUTION
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Part 1: Work done with a horizontal push.

    W = F x d x cos(theta)
    theta = 0 degrees (force and motion are in the same direction)

    W = 80 x 6 x cos(0)
    W = 80 x 6 x 1
    W = 480 J

Part 2: Work done pushing at 30 degrees above horizontal.

    W = F x d x cos(theta)
    theta = 30 degrees

    cos(30 degrees) = 0.866

    W = 80 x 6 x 0.866
    W = 415.7 J (approximately)

ANSWER: The horizontal push does 480 J of work. The angled push does
approximately 415.7 J - LESS work than the horizontal push, for the
same force and the same horizontal distance travelled.

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WHY THIS WORKS AS AN ANSWER

The cos(theta) term in W = Fd cos(theta) measures how much of the
applied force actually points in the direction of motion. At theta =
0 degrees, the entire force acts along the direction of travel, so
cos(0) = 1 and all of it counts as useful work. At theta = 30
degrees, only part of the 80 N force points horizontally (the rest
points partly upward, which does no work against a purely horizontal
displacement), so cos(30 degrees) = 0.866 means only about 86.6% of
the force's magnitude effectively contributes to moving the sofa
horizontally. This is exactly why pushing something at an upward
angle is less efficient than pushing it straight along its direction
of travel - some of the effort is "wasted" lifting rather than moving
it forward.