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John slides down a hill at a 4 m height initially at rest. If all energy is conserved and friction is excluded, what is John's velocity at the bottom of the ramp?

Answer :

ARDNI313GO

The velocity at the bottom : v = 8.85 m/s

Further explanation

Given

height of the hill = 4 m

Required

The velocity at the bottom

Solution

The law of conservation energy :

ME₁=ME₂

(PE+KE)₁ = (PE+KE)₂

initially at rest⇒vo=0⇒KE₁=0

At the bottom⇒h=0⇒PE₂=0

So the equation becomes :

PE₁=KE₂

mgh=1/2.mv²

gh = 1/2v²

[tex]\tt v=\sqrt{2gh}[/tex]

v = √2x9.8x4

v = 8.85 m/s

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