Question

Using conservation of energy, find the speed of the block at the bottom of the ramp. Express your answer in terms of some or all the variables mmm, vvv, and hhh and any appropriate constants.

Answer 1

Answer:

√(v² + 2gh)

Explanation:

PE₀ + KE₀ = PEƒ + KEƒ

At the top of the ramp, potential energy will be at a maximum and kinetic energy will be at a minimum. At the bottom of the ramp, kinetic energy will be at a maximum and potential energy will be minimum. But although the type of energy changes, the total amount of energy remains the same:

mgh + ½mv² = mgh(b) + ½mv(b)²

Since the height at the bottom is zero, we can eliminate potential energy from the right side of the formula:

mgh + ½mv² = ½mv(b)²

gh + ½v² = ½v(b)²

2gh + v² = v(b)²

v(b) = √(v² + 2gh)

== √(v² + 2gh)