Question

A 5.0 kg block moves in a straight line on a horizontal frictionless surface under the influence of a force that varies with position as shown. How much work is done by the force as the block moves from the origin to $x=8.0m$?

• A. 25 J
• B. 50 J
• C. 110 J
• D. 125 J

Answer 1

The correct option is C 110 J

R.E.F image

As force $=$ work. displacement,

in integral form,

$W_{Total}={\int }_a^{b}F.ds$

for displacement from a to b

$\therefore$ W is represented by area under

Force - position curve

Here, it is shown by red shaded area

Area can be calculated as:

$=(20\times 2)+(\frac{1}{2}\times 10\times 2)+(4\times 10)+(5\times 2)+(2\times 5)$

$=(40+10+40+10+10)$ joules

$=110J$ Answer