Byju's Answer

Standard XII

Physics

Work Done as Dot Product

A force F=1...

Question

A force $F = (10 + 0.5 x) N$ acts on a particle in $X$ direction, where $x$ is in meters. Find the work done by this force during a displacement from $x = 0$ to $x = 2$ .

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Solution

Force $F = 10 + 0.5 x$
Work done $W = \int F d x$
$W = \int_{0}^{2} (10 + 0.5 x) d x$
$W = [(10 x + \frac{0.5 x^{2}}{2})]_{0}^{2}$
So, $W = 10 \times (2 - 0) + \frac{0.5}{2} (2^{2} - 0) = 21 J$

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A force F=(10+0.5x)N acts on a particle in X direction, where x is in meters. Find the work done by this force during a displacement from x=0 to x=2.

Force F=10+0.5 xWork done W=∫FdxW=∫20(10+0.5x)dxW=[(10x+0.5x22) ]20So, W=10×(2−0)+0.52(22−0)=21 J

A force $F = (10 + 0.5 x) N$ acts on a particle in $X$ direction, where $x$ is in meters. Find the work done by this force during a displacement from $x = 0$ to $x = 2$ .

Force $F = 10 + 0.5 x$
Work done $W = \int F d x$
$W = \int_{0}^{2} (10 + 0.5 x) d x$
$W = [(10 x + \frac{0.5 x^{2}}{2})]_{0}^{2}$
So, $W = 10 \times (2 - 0) + \frac{0.5}{2} (2^{2} - 0) = 21 J$