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1.3.2 · D5Work, Energy & Power

Question bank — Work done by variable force — integration

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Every answer below gives the reasoning, because in physics a bare "yes/no" is worthless — the why is the physics.

True or false — justify

The area under a Force–position graph equals the work done.
True — the integral is the limit of summing rectangles , and that sum is exactly the geometric area under the – curve.
If the force is constant, you cannot use ; you must use .
False — a constant force is just a special case, ; the integral always works, it merely simplifies to a rectangle.
Area under an – graph that dips below the axis still counts as positive work.
False — area below the axis means (force opposes motion there), so those slivers are negative and subtract from the total work.
Work done to stretch a spring from to equals the work done by the spring over the same stretch.
False — they are equal in magnitude but opposite in sign: you do against the spring, the spring does on the object.
A force always perpendicular to the motion does zero work, no matter how large it is.
True — since at every instant, the total integral is zero regardless of the force's magnitude.
Swapping the limits of leaves the work unchanged.
False — swapping limits flips the sign, which physically means reversing the direction of travel; the work then has the opposite sign.
If a force graph is a straight line from up to over distance , the work is .
False — it's the triangle area ; the force averages to , not , over the trip.
The dot product in can be dropped in one-dimensional problems.
Half-true — in 1-D along the motion , so it reduces to , but you must keep the sign: a backward force still gives .
Integration is only needed when force depends on position .
False — you integrate whenever force isn't constant; if depends on time or velocity you still sum , just recast the integral into the matching variable.

Spot the error

" and the body moves , so ."
The error is using the final force value as if it were constant; force grows from , so you integrate: , half the wrong answer.
"The spring force is , so work by the spring stretching to is ."
Two errors: the spring force is restoring, , so work by the spring is — negative because it pulls back against the outward motion.
" only, so for a force at to the motion, ."
Missing the angle factor; the correct sliver is , so only the component along motion counts — here half the force.
"Force is N over m, area is a triangle of J, so J."
The geometry is right but the sign is dropped — the force is negative throughout, so the area lies below the axis and J.
"The units of are newtons, since force is in newtons."
Wrong — an integral multiplies force by a length element (metres), so the units are (joules), i.e. energy.
"A gravity that weakens with altitude means still works exactly for a rocket."
assumes constant; if gravity varies you must integrate , and is only a low-altitude approximation.

Why questions

Why does chopping the path into tiny slices let us treat a varying force as constant?
Over a vanishingly small step the force has no room to change appreciably, so is an honest rectangle; the error from pretending it's constant shrinks to zero as the slice shrinks.
Why is the integral, and not simple multiplication, the right tool for variable force?
Multiplication needs a single force value, which doesn't exist when changes; the integral is defined as the limit of adding infinitely many tiny pieces, exactly matching "sum of push over each step."
Why does a normal force or centripetal force do zero work even while acting continuously?
Both point perpendicular to the instantaneous displacement, so at every point; each sliver contributes nothing, and the running total stays zero.
Why does the sign of work tell us about energy flow?
Positive work means the force pushes along the motion, feeding energy into the body; negative work means the force opposes motion, draining energy out — the sign is literally the direction of energy transfer.
Why must we say which force we're integrating when quoting ?
Because "work by the spring" and "work by you against the spring" have opposite signs; the same magnitude describes two different energy transfers depending on which force's you summed.
Why can the average-force shortcut replace integration for a spring but not for every force?
Because the spring force is linear in , so its average is simply the midpoint ; for a nonlinear force like the average isn't the midpoint value, and only the integral gets it right.

Edge cases

If a body returns to its start () under a position-only force, what is the total work?
For a conservative force the round trip gives zero work — the integral over a closed path cancels; see Conservative forces and potential energy.
If the displacement is zero but a huge force is applied, how much work is done?
Exactly zero — with every sliver vanishes; work needs motion, not just force (holding a weight still does no physics-work).
What happens to as ?
It goes to zero like (quadratically), so tiny stretches cost negligible work — consistent with the triangle area shrinking to a point.
A force flips sign midway: for the first half of the path, for the second. What does the total work represent?
The net signed area — positive contributions and negative contributions partially cancel, so the total work is the difference, possibly zero if the areas match.
What is the work if the force is zero over the entire path but the body still moves?
Zero work — every sliver is ; a moving body with no force along it (free glide) has no work done on it, so its kinetic energy stays constant by the Work–Energy Theorem.
If is discontinuous (jumps suddenly, like Example 3's kink at m), can we still integrate?
Yes — the area under the curve is still well-defined; we just add the piece before the jump to the piece after (triangle plus rectangle), since a single point of discontinuity contributes no area.

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