A sled is initially given a shove up a frictionless 18.0° incl...

Question

A sled is initially given a shove up a frictionless 18.0° incline. It reaches a maximum vertical height 1.22 m higher than where it started at the bottom. What was its initial speed?

Accepted Answer

Welcome back. Everyone in this problem at a bargain is initially pushed up a friction free slope set at a 17 degree angle, achieving a height of 1.23 m above its launch point, determine its initial speed. For our answer choices is is it's 4.91 m per second. B 5.19 m per second. C 7.19 m per second and D 9.99 m per second. Now, if we're going to figure this out, let's try to visualize what's really going on here. So we're talking about the Toboggan. OK? And for a bargain in our problem, we know that the, the incline is set with the horizontal is 17 degrees. And we also know that it achieves a final height of 1.23 m. So here on our diagram, OK, would be 1.23 m vertical height here and theta would be the angle it is inclined with or the, the, the slopes angle that is, and we have our Tobago which a normal force is acting perpendicular to the surface at the Tobago. OK? And the weight of the Tobago is acting downwards. So we have M multiplied by G, let's let M be equal to the Togo's mass. No, the normal force. OK. Since it's perpendicular to the direction of motion here, that force does no work. Thus, the Togo's mechanical energy is conserved. So let's think about both situations. It says it's pushed up a friction free slope. So initially, we can imagine that our Toboggan would be here in red and at this point its height, let's call Y one equals 0 m. Well, its speed, let's call it V one initially is what we don't know. OK. But we know that at the end, OK. After it now gets to the top of the slope, then its height, let's call that Y two is equal to H and its speed V two is going to be equal to 0 m per second because it's no longer moving. Now, what do we know that can help us to determine the initial speed? Well, let's think about what's going on here. We're talking about movement, we're talking about energy and recall that the principle of energy conservation tells us that the energy energy cannot be destroyed, it can only be changed from one form to another. So that means the initial kinetic energy and potential energy, which would be the two types of energy involved here for our movement must be equal to our final kinetic energy. And our final potential energy recall that kinetic energy in this case, let me call it EK is equal to half MV squared and potential energy is equal to mass multiplied by gravity multiplied by height MG. OK. So in that case, then since we know, let's make a note of what we just said, since we know that the initial kinetic energy plus the initial potential energy is equal to the final kinetic and potential energies, we can substitute what we know to solve. We can use that idea to solve for V one. As we said, that means our initial kinetic energy is going to be half MV one squared K plus initial potential energy mg multiplied by its initial location. Y one and all of that is going to be equal to the final kinetic energy. Half MV two squared plus MG multiplied by Y two. Now remember we said we know that Y one is zero. OK? We know why one is zero. Let me share it here. So we know that Y one is zero and we also know that V two is zero based on our diagram here. So we know that this expression mGy one is going to be zero and half MV two squared is going to be zero. What does that tell us? Well, no, that means that half MV one squared E is really equal to MG by two. We know on both sides, there's a mass so we can cancel our mass which tells us that half V one squared equals G two, we know G the acceleration due to gravity to be 9.8 m per second squared. And we know why to our final height. So let's transpose the sulfur V one because no, this means V one squared is going to be equal to two gy two. And if we square root both sides V one equals the square root of two gy two, when we substitute these values, that means it's going to be the square root of two multiplied by 9.8 m per second squared, multiplied by a final height 1.23 m. If we put this expression into a calculator, we should find that V one or initial speed is 4.909 m per second, which when rounded to two decimal places are three significant figures that is equals 4.91 m per second. Thus the initial speed of the Toon is 4.91 m per second. If we look back at our answer choices, that means A is the correct answer. Thanks for watching everyone. I hope this video helped.

A sled is initially given a shove up a frictionless 18.0° incline. It reaches a maximum vertical height 1.22 m higher than where it started at the bottom. What was its initial speed?

Key Concepts

Conservation of Energy

Gravitational Potential Energy

Kinematics and Initial Velocity

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