Question Video: Mechanical Energy Conversion | Nagwa Question Video: Mechanical Energy Conversion | Nagwa

Question Video: Mechanical Energy Conversion Physics • First Year of Secondary School

Join Nagwa Classes

Attend live Physics sessions on Nagwa Classes to learn more about this topic from an expert teacher!

A skateboarder rolls upward along a curved ramp, as shown in the diagram, and reaches a point that is a vertical height of 7.5 m above the base of the ramp. What is the skateboarder’s speed at the point that is 1.1 m vertically above the base of the ramp? Give your answer to one decimal place.

12:33

Video Transcript

A skateboarder rolls upward along a curved ramp, as shown in the diagram, and reaches a point that is a vertical height of 7.5 meters above the base of the ramp. What is the skateboarder’s speed at the point that is 1.1 meters vertically above the base of the ramp? Give your answer to one decimal place.

Okay, so in this first part of the question, we are asked to find the speed of the skateboarder when they’re at a height of 1.1 meters. We’ll label this height as ℎ one so that we have ℎ one is equal to 1.1 meters. We’ll label the skateboarder’s speed at this height as 𝑣 one, and this is the quantity that we’re trying to find. We are told that the skateboarder reaches a maximum vertical height of 7.5 meters. We’ll call this height ℎ two so that we have ℎ two is equal to 7.5 meters. Since this height of 7.5 meters is the maximum height reached by the skateboarder, we know that at this height the skateboarder is changing direction. They’ve just gone up the ramp and they’re about to start coming back down.

If the skateboarder is changing direction, this means that they’re going from moving in one direction to moving back in the opposite direction. So their speed must instantaneously be zero meters per second at the point at which the direction changes. This change of direction occurs at a height of ℎ two equal to 7.5 meters. So we know that at this point, the speed of the skateboarder is zero meters per second. We’ll label this speed as 𝑣 two so that we have 𝑣 two is equal to zero meters per second. As the skateboarder rose up the ramp, they’re gaining height and so gaining gravitational potential energy. At the same time, the skateboarder’s speed is decreasing from some initial value 𝑣 one to a speed 𝑣 two equals zero meters per second. This means that the skateboarder is losing kinetic energy as they roll up the ramp.

We can recall that the principle of energy conversion and conservation means that the kinetic energy, or 𝐾𝐸, lost by the skateboarder must be equal to the gravitational potential energy, or 𝐺𝑃𝐸, that they gain. We can also recall that the kinetic energy of an object is equal to a half multiplied by 𝑚, the object’s mass, multiplied by the square of 𝑣, the object’s speed. Meanwhile, the gravitational potential energy of an object is equal to 𝑚, the object’s mass, multiplied by 𝑔, the acceleration due to gravity, multiplied by ℎ, the object’s height. On Earth, 𝑔 has a value of 9.8 meters per second squared.

Since the kinetic energy of an object depends on the speed of that object, then the kinetic energy lost by our skateboarder is going to depend on the initial speed 𝑣 one. So let’s work out an expression for the kinetic energy lost by our skateboarder between this point at a height ℎ one and this point at a height ℎ two. The kinetic energy lost must be equal to the initial kinetic energy, which is a half times 𝑚 times 𝑣 one squared, minus the final kinetic energy, a half times 𝑚 times 𝑣 two squared.

Now, in both these terms on the right-hand side, we have a factor of a half and a factor of 𝑚. So we can factor these two terms out. This gives us that the kinetic energy lost is equal to a half 𝑚 multiplied by 𝑣 one squared minus 𝑣 two squared. Since we know that the kinetic energy lost must be equal to the gravitational potential energy gained, let’s also work out an expression for the gravitational potential energy gained by the skateboarder. The gravitational potential energy gained must be equal to the final gravitational potential energy, so that’s 𝑚 times 𝑔 times ℎ two, minus the initial gravitational potential energy. That’s 𝑚 times 𝑔 times ℎ one.

We can notice that there is both an 𝑚 and a 𝑔 in both terms on the right-hand side, which we can factor out. This gives us that the gravitational potential energy gained is equal to 𝑚 times 𝑔 times ℎ two minus ℎ one. So we now have an expression for the kinetic energy lost by the skateboarder and for the gravitational potential energy gained by the skateboarder. The principle of energy conversion and conservation tells us the kinetic energy lost must be equal to the gravitational potential energy gained. This means that we can equate these two expressions. Doing this gives us that a half times 𝑚 times 𝑣 one squared minus 𝑣 two squared is equal to 𝑚 times 𝑔 times ℎ two minus ℎ one. Since the mass 𝑚 appears on both sides of this equation, we can cancel it out. This means that our result will not depend on the mass of the skateboarder.

Then, in this equation, we’re trying to find the value of 𝑣 one. And we know the values of 𝑣 two, ℎ two, and ℎ one. So let’s rearrange the equation to make 𝑣 one the subject. We’ll start by multiplying both sides of the equation by two. On the left-hand side, we have two multiplied by a half, which gives us one. And then, one times 𝑣 one squared minus 𝑣 two squared is simply 𝑣 one squared minus 𝑣 two squared.

Next, we’ll add 𝑣 two squared to both sides of the equation. On the left-hand side, we now have a 𝑣 two squared and a minus 𝑣 two squared. These two terms negate each other. And we’re left with an equation that says that 𝑣 one squared is equal to two times 𝑔 times ℎ two minus ℎ one plus 𝑣 two squared. Taking the square root of both sides of this equation, we get that 𝑣 one is equal to the square root of two times 𝑔 times ℎ two minus ℎ one plus 𝑣 two squared.

Now, we just need to substitute in our values for ℎ one, ℎ two, and 𝑣 two into this equation. So let’s clear some space to do this. Okay, substituting our values into this equation, we get this expression for the speed 𝑣 one. Let’s quickly remind ourselves of where these numbers come from. Looking at this expression, our first term under the square root is two multiplied by 𝑔 multiplied by ℎ two minus ℎ one. Now, looking over here, this first term is two multiplied by 9.8 meters per second squared, so that’s our value of 𝑔, multiplied by 7.5 meters minus 1.1 meters. So that’s the value of ℎ two minus the value of ℎ one.

Looking again at this symbolic expression, we see that the second term under the square root is the square of the speed 𝑣 two. Now, in this case, we know that the speed 𝑣 two is equal to zero meters per second. And so, in this expression for 𝑣 one, the second term under the square root is the square of zero meters per second. Evaluating the expression under the square root, we find that 𝑣 one is equal to the square root of 125.44 meters squared per second squared. Then, evaluating the square root, we find that the speed 𝑣 one is equal to 11.2 meters per second.

We note that this result is exact and has not been rounded. If we look back at the question again, we see that we were asked to give our answer to one decimal place. The result that we have calculated has one decimal place. And so this is our answer to the question that the skateboarder’s speed at the point that is 1.1 meters vertically above the base of the ramp is 11.2 meters per second. Now let’s look at the second part of the question.

What would the initial speed at the base of the ramp required for the skateboarder to reach the top of the ramp be? Give your answer to one decimal place.

As with the first part of the question, we can solve this using the principle of energy conversion and conservation. Specifically, we know that the kinetic energy lost by the skateboarder as they roll up the ramp is equal to the gravitational potential energy that they gain. We’re asked to find the initial speed at the base of the ramp that’s required for the skateboarder to reach the top. We’ll label this initial speed as 𝑣 one. This is the quantity that we’re trying to calculate. 𝑣 one is the skateboarder’s speed at the base of the ramp. And at the base of the ramp, the skateboarder’s height above the ground is going to be zero. We’ll label this height as ℎ one so that we have ℎ one is equal to zero meters.

If the skateboarder just reaches the top of the ramp, then we know that their speed at the top of the ramp is going to be zero meters per second because this is the point at which the skateboarder changes direction. We’ll label this speed as 𝑣 two so that we have 𝑣 two is equal to zero meters per second. We can see from the diagram that the top of the ramp has a height above the ground of 7.5 meters plus 2.7 meters. If we label this height at the top of the ramp as ℎ two, then we have that ℎ two is equal to 7.5 meters plus 2.7 meters. Evaluating this gives us that ℎ two is equal to 10.2 meters.

As with the first part of the question, we’ll work out the kinetic energy lost by the skateboarder and the gravitational potential energy gained by the skateboarder and equate these two in order to calculate the speed 𝑣 one. Recall that kinetic energy is equal to a half times mass times speed squared. And gravitational potential energy is equal to mass times 𝑔 times ℎ, where 𝑔 is the acceleration due to gravity and is equal to 9.8 meters per second squared. The expressions for the kinetic energy lost by the skateboarder and the gravitational potential energy gained are the same as in the first part of the question. So the kinetic energy lost is equal to a half multiplied by the mass 𝑚 multiplied by 𝑣 one squared minus 𝑣 two squared. And the gravitational potential energy gained is equal to the mass 𝑚 multiplied by 𝑔 multiplied by ℎ two minus ℎ one.

As before, the principle of conversion and conservation of energy tells us that we can equate these two expressions. And this gives us this expression here. Just as in the first part of the question, we’re trying to find the speed 𝑣 one. So we need to rearrange this to make 𝑣 one the subject. The method is exactly the same rearrangement as in the first part of the question. And it gives us that 𝑣 one is equal to the square root of two times 𝑔 times ℎ two minus ℎ one plus 𝑣 two squared.

Let’s now clear some space so that we can substitute our values for ℎ one, ℎ two, and 𝑣 two into this expression. Okay, substituting in our values, we have that 𝑔 is equal to 9.8 meters per second squared. Our value for ℎ two is 10.2 meters, our value for ℎ one is zero meters, and our value for the speed 𝑣 two is zero meters per second. This expression then simplifies to give us that 𝑣 one is equal to the square root of two times 9.8 meters per second squared times 10.2 meters. Evaluating the expression under the square root gives a result of 199.92 meters squared per second squared. Then, taking the square root gives us that the speed 𝑣 one is equal to 14.139 meters per second, where the ellipses indicate that there are further decimal places.

The question asks us to give our answer to one decimal place. Rounding this result to one decimal place gives us our answer to the question that, in order to reach the top of the ramp, the speed of the skateboarder at the base of the ramp must be equal to 14.1 meters per second.

Join Nagwa Classes

Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher!

  • Interactive Sessions
  • Chat & Messaging
  • Realistic Exam Questions

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy