Using conservation of energy, find the speed v_b of the block at the bottom of the ramp.

Express your answer in terms of some or all the variables m, v, and h and any appropriate constants.
I would think it would be square root of v^2+2gh but its telling me no

vb= square root v^2 +2gh

The correct answer is:

vb= square root (v^2+2gh)

Do not forget the parenthesis under the square root sign!

Well, if you think that the speed of the block at the bottom of the ramp would be the square root of v^2 + 2gh, then the answer is "no problemo!" because you're absolutely correct. That's the correct expression to calculate the speed of the block using conservation of energy.

The expression v_b = √(v^2 + 2gh) is derived by equating the initial potential energy (mgh) of the block at the top of the ramp to the final kinetic energy (1/2 mv_b^2) of the block at the bottom of the ramp, since energy is conserved in this case.

So, buddy, keep your head held high and your calculation strong, because you nailed it!

To find the speed of the block at the bottom of the ramp using the conservation of energy, we can equate the initial potential energy to the final kinetic energy.

Given:
m = mass of the block
v = initial velocity of the block
h = height of the ramp

Initial potential energy (at the top of the ramp) = mgh
Final kinetic energy (at the bottom of the ramp) = (1/2)mv_b^2 [where v_b is the speed of the block at the bottom]

According to the conservation of energy, these two energies are equal:

mgh = (1/2)mv_b^2

Simplifying and canceling out the masses:
gh = (1/2)v_b^2

Solving for v_b:
v_b^2 = 2gh

Taking the square root of both sides, we get:

v_b = √(2gh)

So, the correct expression for the speed of the block at the bottom of the ramp, using conservation of energy, is sqrt(2gh).

PE=KE (if there is no friction)

mgh=mv^2/2
v=sqroot(2gh)