A $$20$$ kg block attached to a spring of spring constant $$5$$ $$N/m$$ is released from rest at A. The spring at this instant is having an elongation of $$1m$$. The block is allowed to move in smooth horizontal slot with the help of constant force of $$50$$ N in the rope as shown. The velocity of block as it reaches B (in$$ $$m/$$$$sec$$$$) is.

Question

Infinity Learn · Accepted Answer

By the work energy principle  $$wex+wspring=$$ΔK.E     $$50$$×$$2$$−$$(5$$×$$522$$−$$5$$×$$122)=12mV2$$     $$100$$−$$60=12$$×$$20V2$$     $$V=2m/$$$$sec$$

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By the work energy principle
_{$w_{e_{x}} + w_{s p r i n g} = Δ K . E$}

_{$50 \times 2 - (\frac{5 \times 5^{2}}{2} - \frac{5 \times 1^{2}}{2}) = \frac{1}{2} m V^{2}$}

_{$100 - 60 = \frac{1}{2} \times 20 V^{2}$}

_{$V = 2 m / \sec$}

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Work

Remember concepts with our Masterclasses.

By the work energy principle wex+wspring=ΔK.E 50×2−(5×522−5×122)=12mV2 100−60=12×20V2 V=2m/sec

Ready to Test Your Skills?

free study material

By the work energy principle
_{$w_{e_{x}} + w_{s p r i n g} = Δ K . E$}

_{$50 \times 2 - (\frac{5 \times 5^{2}}{2} - \frac{5 \times 1^{2}}{2}) = \frac{1}{2} m V^{2}$}

_{$100 - 60 = \frac{1}{2} \times 20 V^{2}$}

_{$V = 2 m / \sec$}