A body of mass $1 k g$ is moving in a vertical circular path of radius $1 m$ . The difference between the kinetic energies at the highest and lowest point is :

20 J

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10 J

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4 \sqrt{5} J

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10 (\sqrt{5} - 1) J

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Solution

The correct option is A $20 J$
Given,

$r = 1 m$

$m = 1 k g$

$g = 10 m / s^{2}$

In limiting case, for body to revolve in circle,

$V_{t o p} = v_{1} = \sqrt{r g}$

$V_{b o t t} = v_{2} = \sqrt{5 r g}$

The difference in kinetic energy,

$Δ E = (K . E)_{b o t t} - (K . E)_{t o p}$

$Δ = \frac{1}{2} m (v_{2}^{2} - v_{1}^{2})$

$Δ = \frac{1}{2} m (5 r g - r g) = \frac{4 m r g}{2}$

$Δ E = \frac{4 \times 1 \times 1 \times 10}{2}$

$Δ E = 20 J$

The correct option is A.

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A body of mass 1 kg is moving in a vertical circular path of radius 1m. The difference between the kinetic energies at the highest and lowest point is :

The correct option is A 20 JGiven,r=1mm=1kgg=10m/s2In limiting case, for body to revolve in circle,Vtop=v1=√rgVbott=v2=√5rgThe difference in kinetic energy,ΔE=(K.E)bott−(K.E)top Δ=12m(v22−v21)Δ=12m(5rg−rg)=4mrg2ΔE=4×1×1×102ΔE=20JThe correct option is A.

A body of mass $1 k g$ is moving in a vertical circular path of radius $1 m$ . The difference between the kinetic energies at the highest and lowest point is :