Question

A force $F=(10+0.50x)$ acts on a particle in the $x-$direaction. Find the work done by this force during a displacment from $x=0$ to $x=2.0m$

• A. $20J$
• B. $22J$
• C. $21J$
• D. $19J$

Answer 1

The correct option is C $21J$

Given that,

$F=10+0.50x$

Displacement

$x_1=0$

$x_2=2.0m$

We know that,

The work done is

$W=\underset{0}{\overset{2.0}{\int }}F\cdot dx$

$W=\underset{0}{\overset{2.0}{\int }}(10+0.50x)dx$

$W={(10x+0.50\frac{x^2}{2})}_0^{2.0}$

$W=10\times 2.0+0.50\times \frac{2.0\times 2.0}{2}$

$W=21J$

Hence, the work done is $21J$