Variable Force Motion
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Direct answer
The moment force depends on position, time or velocity, the constant-acceleration equations collapse and physics returns to its definition: acceleration is F/m with F evaluated at the instant, work is the area under the force–displacement curve, W = ∫F dx, and velocity follows from ½mv² − ½mu² = ∫F dx. For time-dependent forces, the impulse area ∫F dt = Δp replaces it. Position-dependent forces yield to energy; velocity-dependent forces (like viscous drag F = −kv) yield to separable first-order equations whose signature outcomes are exponential decay and terminal velocity v_t = mg/k. JEE tests exactly this switch — recognising which variable the force depends on and picking area-under-graph or integration accordingly.
What you must remember
- Position-dependent force: use W = ∫ from x1 to x2 of F(x) dx; for F = 3 + 2x newtons from x = 0 to x = 4 m, W = [3x + x²] = 28 J, and v = √(2W/m) from rest.
- F–x graph: the area between the curve and the x-axis is the work; area below the axis counts negative, and JEE regularly supplies the force as a trapezium or triangle sketch instead of a formula.
- Impulse for time-dependent force: ∫F dt = change in momentum; for a force that ramps linearly from 0 to F0 in time t, the impulse is ½F0 t — the average force is half the peak.
- Power connection: instantaneous power P = Fv, so a vehicle at constant power has F = P/v and diminishing acceleration.
- Linear drag F = −kv: v = v0 e^(−kt/m) for a coasting body; velocity falls exponentially with time constant m/k, distance travelled before stopping is finite, mv0/k.
- Gravity plus drag: a falling body with linear drag obeys v = v_t(1 − e^(−gt/v_t)) with terminal velocity v_t = mg/k; at t = m/k the speed is 63 percent of terminal.
- Spring force F = −kx: the archetype of variable force, giving SHM by energy conservation, ½kA² = ½mv_max²; recognise any F ∝ −x law as oscillatory.
Walking through two regimes
A 2 kg particle at rest at the origin feels F = (3 + 2x) N. Work to x = 4 m is the area under the line: ∫(3 + 2x)dx = 3(4) + (4²) = 28 J, so v = √(2 × 28/2) = √28 ≈ 5.3 m/s. No kinematics formula could have produced this — only the area idea. Had the force been given as a graph peaking at 8 N at x = 2 m and falling linearly to zero at x = 4 m, the same answer would come from the trapezium area, which is why JEE Main alternates between formula and sketch for identical physics.
The velocity-dependent case runs differently. A boat of mass m moving at v0 shuts its engine and feels drag F = −kv. Newton's law gives m dv/dt = −kv, so v = v0 e^(−kt/m): the speed never mathematically reaches zero, but the distance to stop is finite, ∫v dt = mv0/k — a result worth memorising because options often include "infinite distance" as the trap. Add gravity for a raindrop-style problem: m dv/dt = mg − kv gives terminal speed mg/k, approached exponentially, and at terminal speed the net force is zero, so the drop thereafter falls at constant velocity through still air. The three-variable classification — F(x), F(t), F(v) — tells you within five seconds which tool the problem needs: energy, impulse, or the exponential integral.
How the exam separates the prepared
JEE Main mostly tests graph literacy: given a force–time triangle, compute the final velocity; given F–x, locate turning points where the area cumulatively returns to zero. JEE Advanced complicates the classification — a force like F = −kv² (quadratic drag) still separates, giving v_t = √(mg/k), while mixing gravity and position-dependence in one problem (a chain falling off a table, where the driven mass itself varies) demands writing F = (m/L) x g and integrating. The standard trap is averaging wrongly: the average of a linearly varying force is (initial + final)/2, but students apply this to non-linear forces. A subtler one is sign — an area below the axis in an F–x plot reduces the net work, and options are usually spaced to catch students who took every area as positive. When force, velocity and displacement directions tangle, fix a sign convention before the first line of algebra.
Frequently asked questions
How do you find velocity from a position-dependent force?
Compute the work W = ∫F dx between the two positions and equate to the change in kinetic energy, ½mv² − ½mu² = W; starting from rest, v = √(2W/m).
What does the area under a force–time graph give?
The impulse, equal to the change in momentum; dividing by the time interval gives the average force, which for a linear ramp is half the peak force.
Why do coasting bodies with drag F = −kv never stop in theory?
The velocity decays as v0 e^(−kt/m), approaching zero only asymptotically, though the total distance covered is the finite value mv0/k.
What is terminal velocity under linear drag?
v_t = mg/k, reached when drag balances weight; the approach is exponential with time constant m/k, at which the speed is about 63 percent of terminal.
When must you abandon the constant-acceleration equations?
The instant acceleration depends on position, time or velocity — which is any force of the form F(x), F(t) or F(v) — since the SUVAT set assumes constant a throughout.