A block of mass 4 kg is at rest on a frictionless horizontal surface. A horizontal force as a function of displacement as shown in graph acts on block. Speed of block when it has displaced by 6m is (Write two digits after the decimal point.)

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Solution

By work energy theorem,
Work $= Δ K . E$
Work = Area under $F - x$ graph.
So work $= \frac{1}{2} \times 2 \times 2 + 2 \times 2 + 2 \times 4 + \frac{1}{2} \times 2 \times 4 = 18 J$
$Δ K . E = \frac{1}{2} m v^{2} = 2 v^{2}$
Work $= Δ K . E$
$18 = 2 v^{2} \Rightarrow v = 3 m / s$

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A block of mass 4 kg is at rest on a frictionless horizontal surface. A horizontal force as a function of displacement as shown in graph acts on block. Speed of block when it has displaced by 6m is (Write two digits after the decimal point.)

By work energy theorem, Work =ΔK.E Work = Area under F−x graph. So work =12×2×2+2×2+2×4+12×2×4=18 J ΔK.E=12mv2=2v2 Work =ΔK.E 18=2v2⇒v=3m/s

By work energy theorem,
Work $= Δ K . E$
Work = Area under $F - x$ graph.
So work $= \frac{1}{2} \times 2 \times 2 + 2 \times 2 + 2 \times 4 + \frac{1}{2} \times 2 \times 4 = 18 J$
$Δ K . E = \frac{1}{2} m v^{2} = 2 v^{2}$
Work $= Δ K . E$
$18 = 2 v^{2} \Rightarrow v = 3 m / s$