A block slides down a ramp from rest. The block travels 7.19 m/s at the bottom of the ramp and has a mass of 2.98 kg. Assuming no friction, what is the height of the block off the ground at the top of the ramp?
As we assume that there is no friction, we can apply the law of conservation of energy to two different moments of the block's movement:
The energy at the top is "E_{kinetic}+E_{potential}=mgH", as the velocity at the top is zero (block is initially at rest)
The energy at the bottom is "E_{kinetic}+E_{potential}=\\frac{mv^2}{2}", the block has descended, but gained a velocity.
Thus, by the energy conservation, "\\frac{mv^2}{2}=mgH; H=\\frac{v^2}{2g}; H\\approx 2.63m" (where i took "g=9.8 m\/s^2").
Comments
Leave a comment