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Question Video: Comparing Changes in Height for an Object with Constant Mechanical Energy Physics • 9th Grade

The velocity and height above ground level, ℎ₀, of an object is shown at different times in the figure. The figure is not to scale. The mechanical energy of the object is constant. Which of the following has the greatest magnitude? [A] ℎ₁ − ℎ₀ [B] ℎ₂ − ℎ₁ [C] ℎ₃ − ℎ₂ [D] ℎ₄ − ℎ₁

06:09

Video Transcript

The velocity and height above ground level, ℎ zero, of an object is shown at different times in the following figure. The figure is not to scale. The mechanical energy of the object is constant. Which of the following has the greatest magnitude? (A) ℎ sub one minus ℎ sub zero. (B) ℎ sub two minus ℎ sub one. (C) ℎ sub three minus ℎ sub two. (D) ℎ sub four minus ℎ sub one.

In this question, we are given a figure showing the motion of an object as it travels along a curve. We are given the velocity of the object at various points along its journey. And we want to find the height of the object at each interval its velocity was measured at to determine which of the answer options have the greatest magnitude. Before we begin to solve this, we need to recall some information about mechanical energy and how we can use it to calculate the values that we want.

We can recall that the mechanical energy, ME, of an object is defined as the sum of its kinetic energy and potential energy. The kinetic energy, KE, is equal to one-half multiplied by the mass, 𝑚, multiplied by the velocity, 𝑣, squared. We also have gravitational potential energy, GPE, in this problem which is equal to the mass, 𝑚, multiplied by the acceleration due to gravity, 𝑔, multiplied by the height of the object, ℎ. Therefore, the mechanical energy of the object is equal to one-half 𝑚𝑣 squared plus 𝑚𝑔ℎ.

Because we are told the mechanical energy for the object is constant throughout its travel, we can set the mechanical energy at each of these points to be equal to each other to find the difference in height. So for the general case of two different points, which we will label as point A and point B, we will have the equation one-half 𝑚 𝑣 sub A squared plus 𝑚𝑔 ℎ sub A equals one-half 𝑚 𝑣 sub B squared plus 𝑚𝑔 ℎ sub B. We now want to rearrange this so that we have an equation for the difference in heights ℎ sub B and ℎ sub A.

We can do this by subtracting 𝑚𝑔 ℎ sub A from both sides and subtracting one-half 𝑚 𝑣 sub B squared from both sides to give us one-half 𝑚 𝑣 sub A squared minus one-half 𝑚 𝑣 sub B squared equals 𝑚𝑔 ℎ sub B minus 𝑚𝑔 ℎ sub A. If we now factorize both sides, this will give us one-half 𝑚 multiplied by 𝑣 sub A squared minus 𝑣 sub B squared equals 𝑚𝑔 multiplied by ℎ sub B minus ℎ sub A. We can now divide both sides by 𝑚𝑔, and we will have ℎ sub B minus ℎ sub A equals 𝑣 sub A squared minus 𝑣 sub B squared over two 𝑔.

So we now have a general equation to calculate the difference in heights of this object. Let’s begin with option (A), ℎ sub one minus ℎ sub zero. ℎ sub one minus ℎ sub zero will be equal to 𝑣 sub zero squared minus 𝑣 sub one squared over two 𝑔. From the figure, we can see that 𝑣 sub zero is equal to 20 meters per second, and 𝑣 sub one is equal to 12 meters per second. We can also recall that the acceleration due to gravity 𝑔 is equal to 9.8 meters per second squared. Substituting these values into our equation, we find that ℎ sub one minus ℎ sub zero is equal to 20 meters per second squared minus 12 meters per second squared over two multiplied by 9.8 meters per second squared, which equals 13.1 meters.

Now we can repeat this method for the other three options. For option (B), ℎ sub two minus ℎ sub one will be equal to 𝑣 sub one squared minus 𝑣 sub two squared over two 𝑔. From the figure, we can see that 𝑣 sub one is equal to 12 meters per second, and 𝑣 sub two is equal to 5.6 meters per second. Substituting these values into our equation, we find that ℎ sub two minus ℎ sub one is equal to 12 meters per second squared minus 5.6 meters per second squared over two multiplied by 9.8 meters per second squared, which equals 5.7 meters. We can see that this value is smaller than option (A), so we can eliminate option (B).

For option (C), ℎ sub three minus ℎ sub two will be equal to 𝑣 sub two squared minus 𝑣 sub three squared over two 𝑔. From the figure, we can see that 𝑣 sub two is equal to 5.6 meters per second, and 𝑣 sub three is equal to 1.5 meters per second. Substituting these values into our equation, we find that ℎ sub three minus ℎ sub two is equal to 5.6 meters per second squared minus 1.5 meters per second squared over two multiplied by 9.8 meters per second squared, which equals 1.5 meters. We can see that this value is smaller than option (A), so we can eliminate option (C).

For option (D), ℎ sub four minus ℎ sub one will be equal to 𝑣 sub one squared minus 𝑣 sub four squared over two 𝑔. From the figure, we can see that 𝑣 sub one is equal to 12 meters per second, and 𝑣 sub four is equal to 3.1 meters per second. Substituting these values into our equation, we find that ℎ sub four minus ℎ sub one is equal to 12 meters per second squared minus 3.1 meters per second squared over two multiplied by 9.8 meters per second squared, which equals 6.9 meters. We can see that this value is smaller than option (A), so we can eliminate option (D).

And therefore, option (A) must be the correct answer. ℎ sub one minus ℎ sub zero has the greatest magnitude.

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