A very slippery block of ice slides down a smooth ramp tilted at ...

Question

A very slippery block of ice slides down a smooth ramp tilted at angle θ. The ice is released from rest at vertical height h above the bottom of the ramp. Find an expression for the speed of the ice at the bottom.

Accepted Answer

Everyone in this problem. One end of a frictionless board of length, L is raised to an elevation of E above the ground. The other end is in contact with the ground. The setup forms a ramp. The ramp makes an angle of alpha with the horizontal, an object at the top of the ramp is released from rest and were asked to write the formula that describes the speed of the object when it touches the ground were given four answer choices. Choice A is the square root of two multiplied by E divided by G B is the square root of two multiplied by G multiplied by E see the square root of two multiplied by E divided by G multiplied by sine alpha or D, the square root of G multiplied by sine alpha divided by two multiplied by E. Alright, so let's draw a little diagram of what we have going on here. So we have a frictionless board of length L that's raised to an elevation of E. We have this board with the length of L and it's raised to an elevation of E. So the distance between the end top of that board and the ground is e okay and it makes an angle of alpha with the horizontal. Now, we have an object at the top of the ramp, we're gonna draw it as a ball but any object And it is initially at rest, it is released from rest. So the initial speed of this object v naught is 0m/s, then it's going to be released and it's gonna travel down this ramp. So we have a problem where we have some this motion and we also have some gravity because the object is at some height. So we want to think about conservation of mechanical energy. And we know we have conservation of mechanical energy which tells us that the initial kinetic energy lost the initial potential energy. It's going to be equal to the final kinetic energy plus the final potential energy recall that the kinetic energy is given by one half multiplied by M multiplied by V squared. And in terms of potential energy, we have gravitational potential energy here and the gravitational potential energy is given by the mass and multiplied by the gravitational acceleration G multiplied by the height. So we have that on the left hand side for the initial case where we're using V, not an H not. And on the right hand side where they have the same equations, but we're going to use V F and H F. Now, let's think about this. The initial speed is zero. Okay. It's released from rest. We've written that in our diagram. So we know that the first term on the left hand side, one half multiplied by M multiplied by Bennett squared is going to be zero when the object reaches the ground. Okay, the height is going to be zero. We're gonna take the ground to be kind of our reference points, the height of the ground is going to be zero. And so there's gonna be no gravitational potential energy at the end. Okay. So we're converting all of the gravitational potential energy from the beginning into kinetic energy at the ground. Now, we can divide both sides by the mass M as well. So that will cancel. And we're left with the G each not Is equal to 1/2 V F squared. Now our initial height H all this frictionless board is at an elevation of E. So our initial height is going to be E. So if G multiplied by E is equal to one half multiplied by V F squared, we're looking for that final speed. So we want to isolate for via, we can write that V F squared is equal to, we multiply it by two. So we get two multiplied by G multiplied by E. And the last step here is just to take the square root, we get that V F is equal to the square root of two multiplied by G multiplied by E. Alright. So using our conservation of mechanical energy. We found that the speed of the object when it touches the ground is the square root of two multiplied by G multiplied by E which corresponds with answer choice. B thanks everyone for watching. I hope this video helped you in the next one.

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A very slippery block of ice slides down a smooth ramp tilted at angle θ. The ice is released from rest at vertical height h above the bottom of the ramp. Find an expression for the speed of the ice at the bottom.

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