Consider an elliptically shaped rail PQ in the vertical plane with OP =6 m and OQ =8 m. A block of mass 1 kg is pulled along the rail from P to Q with a force of 15 N, which is always parallel to line P Q (see the figure given). Assuming no frictional losses, the kinetic energy of the block when it reaches Q is (n × 10) joules. The value of n is (Take acceleration due to gravity =10 ms -2 )

Question

Consider an elliptically shaped rail PQ in the vertical plane with OP =6 m and OQ =8 m. A block of mass 1 kg is pulled along the rail from P to Q with a force of 15 N, which is always parallel to line P Q (see the figure given). Assuming no frictional losses, the kinetic energy of the block when it reaches Q is (n × 10) joules. The value of n is (Take acceleration due to gravity =10 ms -2 ) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Using work energy theorem W mg + W F =Δ K . E . - mgh + Fd =Δ K . E . ∴ -1 × 10 × 8+15(10)=Δ KE ∴ Δ K . E .=70 ⇒ n =7

Using work energy theorem Wmg​+WF​=ΔK.E. −mgh+Fd=ΔK.E. ∴−1×10×8+15(10)=ΔKE ∴ΔK.E.=70 ⇒n=7

Using work energy theorem
$W_{m g} + W_{F} = Δ K . E .$
$- m g h + F d = Δ K . E .$
$∴ - 1 \times 10 \times 8 + 15 (10) = Δ K E$
$∴ Δ K . E . = 70$
$\Rightarrow n = 7$