Question

A force on a particle as the function of displacement x (in x-direction) is given by
F=10+0.5x
The work done corresponding to displacement of particle from x=0 to x=2 unit is
A. 25 J
B. 29 J
C. 21 J
D. 18 J

Answer 1

To solve this problem, use the formula for work done due to small displacement by force F. Substitute the values and put the limits of displacement from x=0 to x=2. Solve the equation and find the value of work done. This obtained value will be the The work done corresponding to displacement of particles from x=0 to x=2 units.

Formula used:
$dW= F. dx$

Complete step by step answer:
Given: Force on a particle, F= 10+0.5x
Work done due to small displacement dx by force F is given by,
$dW= F. dx$ ...(1)
Where, dx is a small displacement
Integrating above equation we get,
$W= \int F. dx$ ...(2)
Displacement of the particle is from x=0 to x=2 units. So, equation. (2) becomes,
$W= \int _{0}^{2} (10+0.5x) dx$
$\Rightarrow W= \int _{0}^{2}(10x + \dfrac {0.5{x}^{2}}{2})$
$\Rightarrow W= 10(2) + \dfrac{0.5 \times {2}^{2}}{2}$
$\Rightarrow W= 20 + \dfrac {0.5 \times 4}{2}$
$\Rightarrow W= 20 + \dfrac {2}{2}$
$\Rightarrow W= 20 +1$
$\therefore W= 21 J$
Hence, the work done corresponding to displacement of particles from x=0 to x=2 units is 21 J.

So, the correct answer is “Option C”.

Note:
Students must take care while taking the integration. This is the most common mistake made by the students while solving these types of problems. Students should take care while putting in the limits. If we make even a small mistake then the solution will give us a different solution altogether.