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What would be the speed of a cart at the bottom of a very long frictionless inclined plane if it was released from rest at a height of 10.0 m above the bottom of the track? Assume the bottom of the track has height = 0. Show your work.

User Oli C
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1 Answer

1 vote

Answer:

Step-by-step explanation:

Given

Height from which cart is released is
h=10\ m

as we know energy is Conserved and changes only its form so Potential Energy of Cart converted to kinetic Energy at bottom

Energy at Top
E_T=mgh

where m=mass of cart

g=acceleration due to gravity

h=height of track w.r.t bottom

Energy at bottom
E_B=(1)/(2)mv^2


E_T=E_B


mgh=(1)/(2)mv^2

cancel out common terms


v=√(2* g* h)


v=√(2* 9.8* 10)


v=14\ m/s

User ChoNuff
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