A particle of mass 2kg moves on a smooth horizontal plane under the action of a single force F→=(30i⌢+40j^)N . Under this force it is displaced from (0,0) to (100m,100m). If the initial speed of the particle is 3000m/s then find out its final speed in m/s.

W=F→.r→ =[30i→+40j→].[100i¯+100j¯] = 7000 J W=ΔKE=Kf−Ki 7000=12×2(V2(3000)2) V2=10,000V=100 m/s

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A particle of mass 2kg moves on a smooth horizontal plane under the action of a single force $\vec{F} = (30 \overset{⌢}{i} + 40 \hat{j}) N$ . Under this force it is displaced from (0,0) to (100m,100m). If the initial speed of the particle is $\sqrt{3000} m / s$ then find out its final speed in m/s.

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Detailed Solution

$W = \vec{F} . \vec{r}$
$= [30 \vec{i} + 40 \vec{j}] . [100 \bar{i} + 100 \bar{j}]$
= 7000 J
$W = Δ K E = K_{f} - K_{i}$
$7000 = \frac{1}{2} \times 2 (V^{2} {(\sqrt{3000})}^{2})$
$V^{2} = 10, 000$
V=100 m/s

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