A force of

Question:

A force of $\mathrm{F}=(5 \mathrm{y}+20) \hat{\mathrm{j}} \mathrm{N}$ acts on a particle. The workdone by this force when the particle is moved from $\mathrm{y}=0 \mathrm{~m}$ to $\mathrm{y}=10 \mathrm{~m}$ is

Solution:

$\mathrm{F}=(5 \mathrm{y}+20) \hat{\mathrm{j}}$

$\omega=\int F d y=\int_{0}^{10}(5 y+20) d y$

$=\left(\frac{5 y^{2}}{2}+20 y\right)_{0}^{10}$

$=\frac{5}{2} \times 100+20 \times 10$

$=250+200=450 \mathrm{~J}$

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