Formula For Work Done Calculation by Force Displacement Graph

Physics Formulas

Work Done Calculation by Force Displacement Graph

Let a body, whose initial position is xt, is acted upon by a variable force (whose magnitude is changing continuously) and consequently the body acquires its final position xf

Let bar be the average value of variable force within the interval dx from position x to (x +dx) i.e. for small displacement dx. The work done will be the area of the shaded strip of width dx. The work done on the body in displacing it from position xt to xf will be qual to the sum of areas of all the such strips



w = Area under curve between xi and xf i.e Area under force displacement curve with proper algebraic sign represents work done by the force.

Displacement Equations for these Calculations:

displacement formula


s = displacement

v = final velocity

u = initial velocity

t = time

Different resources always use slightly different variables so you might also encounter this same equation with vi or v0 representing initial velocity (u) and vf representing final velocity (v) given below:

final velocity


s = displacement

vf = final velocity

vi = initial velocity

t = time

Examples For Work Done Calculation by Force Displacement Graph

Problem 1: A 10 kg mass moves along x-axis. Its acceleration as a function of its position is shown in the figure. What is the total work done on the mass by the force as the mass moves from x = 0 to x = 8 cm

a. 8 × 10-2 J

b. 16 × 10-2 J

c. 4 × 10-4 J

d. 1.6 × 10-3 J


Solution: Correct option is a

Work done on the mass = mass × covered area between the graph and displacement axis on a-t graph.

numerical value

Problem 2: The relationship between force and position is shown in the figure given (in one dimensional case). The work done by the force in displacing a body from x – 1 cm to x – 5 cm is

a. 20 ergs

b. 60 ergs

c. 70 ergs

d. 700 ergs

force dyne

Solution: Correct option is a

Work done = Covered area on force-displacement graph = 1 × 10 + 1 × 20 -1 × 20 + 1 × 10 = 20 erg.

Problem 3: The graph between the resistive force F acting on a body and the distance covered by the body is shown in the figure. The mass of the body is 25 kg and initial velocity is 2 m/s. When the distance covered by the body is 5m, its kinetic energy would be

(a) 50 J

(b) 40 J

(c) 20 J

(d) 10 J


Solution: Correct option is (d)

Initial kinetic energy of the body

initial kinetic energy

Final kinetic energy = Initial energy – work done against resistive force (Area between graph and displacement axis)

kinetic value

For More Physics formulas vist main page and do solve exercise of NCERT from Physics Wallah.  NCERT solutions for class 11 Physics and NCERT solutions for class 12 Physics.

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