MCQ
The 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
- ✓$21$
- B$29$
- C$18$
- D$25$
✓
Answer
Correct option: A.
$21$
a
To determine the overall work done, we calculate the small work done for the displacement $d x$ as
To determine the overall work done, we calculate the small work done for the displacement $d x$ as
$d W=F \cdot d x=F d x \cos \theta$
$=F d x$
$\left(\because \cos \theta=\cos 0^{\circ}=1\right)$
Now, the total work done for the displacement from $x=0$ to $x=2$ unit
$\int_0^W d W=\int_0^2 F d x=\int_0^2(10+0.5 x) d x$
$=\left[10 x+0.5 \frac{x^2}{2}\right]_0^2=21 J$
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