Consider an elliptical shaped rail in the vertical plane with and . A block of mass 1 kg is pulled along the rail from P to Q with a force of 18 N, which is always parallel to line PQ (see the figure given). Assuming no frictional losses, the kinetic en

Question

Consider an elliptical shaped rail $P Q$ in the vertical plane with $O P = 3 m$ and $O Q = 4 m$ . A block of mass 1 kg is pulled along the rail from P to Q with a force of 18 N, which is always parallel to line PQ (see the figure given). Assuming no frictional losses, the kinetic energy of the block when it reaches Q is $(n \times 10)$ joules. The value of $n$ is (take acceleration due to gravity $= 10 {ms}^{- 2}$ ) [2014]

Smart Achievers · Accepted Answer

(5)

Work done = Increase in P.E. + gain in K.E.

$O P = 3 m$

$18 \times 5 = 1 \times 10 \times 4 + gain in K.E.$

$∴$ $Gain in K.E. = 50 J = 10 n \Rightarrow n = 5$

Solution

Ans.
(5)

Work done = Increase in P.E. + gain in K.E.

$F \times d = m g h + gain in K.E.$

$18 \times 5 = 1 \times 10 \times 4 + gain in K.E.$

$∴$ $Gain in K.E. = 50 J = 10 n \Rightarrow n = 5$

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Solution

Ans. (5) Work done = Increase in P.E. + gain in K.E. F×d=mgh+gain in K.E. 18×5=1×10×4+gain in K.E. ∴ Gain in K.E.=50 J=10n ⇒ n=5

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Ans.
(5)

Work done = Increase in P.E. + gain in K.E.

$F \times d = m g h + gain in K.E.$

$18 \times 5 = 1 \times 10 \times 4 + gain in K.E.$

$∴$ $Gain in K.E. = 50 J = 10 n \Rightarrow n = 5$