Question

A force F = a + bx acts on a particle in the x-direction, where a and b are constants. Find the work done by this force during a displacement from x = 0 to x = d.
Concept of Physics - 1 , HC VERMA , Chapter " Work and Energy"

Answer 1

Hello Dear.

Here displacement of the particle (S)
= d-0 =d

Now, we have to find the average force.
→Force at x =0
F = a

→Force at x = d
F' = a + bd

∴ Average Force =  $\frac{F + F'}{2}$ 
 Average force = {a + (a + bd) } /2

= a +  $\frac{1}{2} bd$ 

Now we know that , work done = F . S

w = { a + $\frac{1}{2} bd$  }. d

Hence the work done is [  $a + \frac{1}{2} bd$ ] d .


Hope it Helps.

Answer 2

The correct answer is, W=ad+(bd^2)/2

Look at the attachment for explaination.