Question

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.

Answer 1

Answer:

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 $E_B=\frac{1}{2}mv^2$

$E_T=E_B$

$mgh=\frac{1}{2}mv^2$

cancel out common terms

$v=\sqrt{2\times g\times h}$

$v=\sqrt{2\times 9.8\times 10}$

$v=14\ m/s$