Question

A force $F=(2+x)$ acts on a particle in x-direction where $F$ is in newton and $x$ in meter. Find the work done by this force during a displacement from $x=1.0$ m to $x=2.0m$.

• A. $3.5J$
• B. $7J$
• C. $4.5J$
• D. $5.5J$

Answer 1

Answer: (A)

$3.5J$

As the force is variable, we shall find the work done in a small displacement from $x$ to $x+dx$ and then integrate it to find the total work. The work done in this small displacement is

$dW=Fdx=(2+x)dx$

Thus, $W={\int }_{1.0}^{2.0}dW={\int }_{1.0}^{2.0}(2+x)dx$

$={[2x+\frac{x^2}{2}]}_{1.0}^{2.0}=3.5J$