# 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 x_{t}, is acted upon by a variable force (whose magnitude is changing continuously) and consequently the body acquires its final position x_{f}.

Let 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 x_{t} to x_{f} will be qual to the sum of areas of all the such strips

w = Area under curve between x_{i} and x_{f} i.e Area under force displacement curve with proper algebraic sign represents work done by the **force**.

### Displacement Equations for these Calculations:

Where,

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:

Where,

s = displacement

v_{f} = final velocity

v_{i} = 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.

**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

**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

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

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**.