কাজ

The force on a particle as the function of displacement x(in x-direction) is given by F=10+0.5xF=10+0.5x

The work done corresponding to displacement of particle from x=0x=0 to x=2x=2 unit is?

হানি নাটস

Given that F=10+0.5xF=10+0.5x

Work done corresponding to the particle from x=0x=0 to x=2x=2 unit is, W=02F.dx=02(10+0.5x)dx=[10x+0.5x22]02=20+1=21JW= \int_{0}^{2} F.dx= \int_{0}^{2} (10+0.5x)dx= [10x+ \dfrac{0.5x^2}{2}]_0^2= 20+1= 21 J

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