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D2.4 · Investigate transformations between gravitational and kinetic energy

Learn to investigate transformations between gravitational and kinetic energy through clear examples and targeted practice.

Ontario Grade 11 Physics

Energy and Society

Ontario Grade 11 Physics — D2.4

A ball held above the floor can move when released. As it falls, its gravitational energy decreases while its kinetic energy increases. Kinetic energy is energy of motion. In this lesson, the system is the moving object and Earth together. We use the floor as a height reference and choose upward as the positive vertical direction. Energy is a scalar: it has an amount but no direction. Velocity is a vector: it has both an amount and a direction. This distinction matters because the object's direction of motion does not make its kinetic energy negative.

What you will learn

1. Prerequisite bridge: height, mass, and speed

Mass measures how much matter an object contains. Its SI unit is the kilogram, written as kg\mathrm{kg}. Height is a vertical distance measured from a chosen reference level. Its SI unit is the metre, written as m\mathrm{m}. Speed tells how fast an object moves and is measured in metres per second, written as m/s\mathrm{m/s}. Velocity includes speed and direction.
A scalar has magnitude only. Mass, height, speed, and energy are scalars. A vector has magnitude and direction. For example, a velocity of 3.0 m/s3.0\ \mathrm{m/s} downward is not the same vector as 3.0 m/s3.0\ \mathrm{m/s} upward. Both have the same speed, so they have the same kinetic energy.
We set the floor at height zero and call upward positive. An object above the floor has positive height. While it falls, its vertical velocity points downward, opposite to the positive direction. The height reference affects the numerical value of gravitational energy, but changes in gravitational energy depend on the change in height.
The sign convention is +y+y upward, with h=0h=0 at the floor. This gives a clear direction for describing motion and a consistent reference for comparing heights.

2. The energy model and the transformation

Gravitational energy is energy an object has because of its height in a gravitational field. Near Earth's surface, calculate it with Eg=mghE_g=mgh. Here, mm is mass in kilograms, gg is the gravitational field strength, and hh is height above the chosen reference. Use g=9.8 N/kgg=9.8\ \mathrm{N/kg} in these examples. The joule, written as J\mathrm{J}, is the SI unit of energy.
Kinetic energy is energy of motion. Calculate it with Ek=12mv2E_k=\frac{1}{2}mv^2, where vv is speed in metres per second. Since speed is squared, kinetic energy is never negative. A change in direction can change velocity without changing speed or kinetic energy.
For an object moving between two heights, a decrease in gravitational energy can become an increase in kinetic energy. If other transfers are small, the total of these two energies stays nearly constant. This is the simple model used here. Real objects may also transfer some energy to their surroundings through effects such as sound or rubbing against a surface, so the measured kinetic-energy increase may be smaller than the predicted gravitational-energy decrease.
For a fall from rest, the starting kinetic energy is zero. In the ideal model, the loss in gravitational energy equals the gain in kinetic energy. For an object moving upward, kinetic energy can decrease as gravitational energy increases. Energy amounts are scalars, so compare the initial and final amounts rather than assigning a direction to energy.
Eg=mgh,Ek=12mv2E_g=mgh,\quad E_k=\frac{1}{2}mv^2

3. Investigating the transformation

An investigation can test whether a drop's loss of gravitational energy is close to its gain in kinetic energy. A possible procedure is to release an object from rest at a measured height and determine its speed at a lower point. A photogate or video analysis could be used to estimate speed. Measure the mass with a balance and measure vertical heights from the same reference level.
This is a proposed procedure, not a report of completed measurements. In an actual investigation, recorded heights, masses, and speeds would be measured evidence. A prediction from the equations, or a result from a simulation, is not itself experimental measurement.
For each trial, calculate the change in gravitational energy and the change in kinetic energy. Compare the size of the gravitational-energy decrease with the kinetic-energy increase. Repeat trials and consider measurement uncertainty, meaning the limited precision of measuring tools and methods. If the changes do not match closely, check the height reference, speed estimate, release conditions, and possible energy transfers to the surroundings.
Keep the physical system in mind: the object and Earth are included when discussing their gravitational energy. If the investigation tracks the object's motion, record the object's speed and height. Use the same height reference throughout so that the comparison is meaningful.
ΔEg=Eg,f−Eg,i,ΔEk=Ek,f−Ek,i\Delta E_g=E_{g,f}-E_{g,i},\quad \Delta E_k=E_{k,f}-E_{k,i}

4. Reading the result

A useful comparison is between the amount of gravitational energy lost and the amount of kinetic energy gained. During a fall, gravitational energy decreases, so its change is negative under the final-minus-initial convention. Kinetic energy usually increases, so its change is positive. Their signed changes can therefore have opposite signs even when the amounts transferred are similar.
Check that the result makes physical sense. An object falling through a greater vertical distance has a larger decrease in gravitational energy. A faster object has more kinetic energy, and doubling its speed makes its kinetic energy four times as large if its mass stays the same. Always report a direction when reporting velocity, but report speed in the kinetic-energy calculation.
ΔEg+ΔEk≈0\Delta E_g+\Delta E_k\approx 0

Worked example

A released ball gains kinetic energy

A 0.40 kg0.40\ \mathrm{kg} ball is released from rest at 2.0 m2.0\ \mathrm{m} above the floor. Find its predicted speed just before reaching the floor using the simple energy model.
  1. Define the system and direction
    The system is the ball and Earth. The floor is the zero-height reference, and upward is positive. The ball's final velocity is downward, but its speed is positive. Known values are mass 0.40 kg0.40\ \mathrm{kg}, initial height 2.0 m2.0\ \mathrm{m}, final height 0 m0\ \mathrm{m}, and initial speed 0 m/s0\ \mathrm{m/s}. The unknown is final speed.
  2. Apply the energy relationship
    Assume the gravitational-energy decrease becomes kinetic-energy increase. The initial kinetic energy is zero because the ball starts from rest.
    mghi=12mvf2mgh_i=\frac{1}{2}mv_f^2
  3. Substitute and solve
    The mass appears on both sides and cancels. Substitute the height and the course value of gravitational field strength, keeping units in the calculation.
    vf=2(9.8 N/kg)(2.0 m)=6.3 m/sv_f=\sqrt{2(9.8\ \mathrm{N/kg})(2.0\ \mathrm{m})}=6.3\ \mathrm{m/s}
Answer: The predicted final velocity is 6.3 m/s6.3\ \mathrm{m/s} downward. The speed is 6.3 m/s6.3\ \mathrm{m/s}.
Check: The expression inside the square root has units of m2/s2\mathrm{m^2/s^2}, so the answer has units of m/s\mathrm{m/s}. The speed is reasonable for a fall of 2.0 m2.0\ \mathrm{m} and is reported to two significant figures.

Worked example

Comparing energy changes for a falling object

For a hypothetical set of supplied values, a 0.25 kg0.25\ \mathrm{kg} object falls from 1.8 m1.8\ \mathrm{m} to 0.60 m0.60\ \mathrm{m} above the floor. Its stated speed at the lower point is 4.7 m/s4.7\ \mathrm{m/s} downward. Compare the gravitational-energy decrease with the kinetic-energy increase if it began at rest.
  1. Set the system and reference
    The system is the object and Earth. Upward is positive, and the floor is the height reference. The mass is 0.25 kg0.25\ \mathrm{kg}; the initial and final heights are 1.8 m1.8\ \mathrm{m} and 0.60 m0.60\ \mathrm{m}. Initial speed is zero. The supplied final speed is 4.7 m/s4.7\ \mathrm{m/s} downward; these values are hypothetical, not reported measurements.
  2. Calculate the gravitational-energy change
    Use final minus initial. A negative result means gravitational energy decreased as the object moved to a lower height.
    ΔEg=(0.25 kg)(9.8 N/kg)(0.60−1.8 m)=−2.9 J\Delta E_g=(0.25\ \mathrm{kg})(9.8\ \mathrm{N/kg})(0.60-1.8\ \mathrm{m})=-2.9\ \mathrm{J}
  3. Calculate the kinetic-energy change
    The starting kinetic energy is zero. Use speed, not signed velocity, in the kinetic-energy relationship.
    ΔEk=12(0.25 kg)(4.7 m/s)2−0 J=2.8 J\Delta E_k=\frac{1}{2}(0.25\ \mathrm{kg})(4.7\ \mathrm{m/s})^2-0\ \mathrm{J}=2.8\ \mathrm{J}
Answer: The gravitational-energy decrease is 2.9 J2.9\ \mathrm{J} in amount, and the kinetic-energy increase is 2.8 J2.8\ \mathrm{J}. They are close.
Check: Both changes are in joules. The signs are sensible: gravitational energy falls and kinetic energy rises. The small difference could result from measurement limits or energy transferred to the surroundings. The values are rounded to two significant figures.

Worked example

Kinetic energy changes into gravitational energy

A 0.50 kg0.50\ \mathrm{kg} object moves upward at 5.0 m/s5.0\ \mathrm{m/s} from a height of 0.40 m0.40\ \mathrm{m}. Assuming its kinetic energy changes into gravitational energy, estimate its greatest height above the floor.
  1. Define the system and known values
    The system is the object and Earth. The floor is the zero-height reference, with upward positive. The object's initial velocity is upward, so its speed is 5.0 m/s5.0\ \mathrm{m/s}. Its mass is 0.50 kg0.50\ \mathrm{kg} and initial height is 0.40 m0.40\ \mathrm{m}. At its greatest height, its speed is momentarily zero.
  2. Relate the energy amounts
    At the greatest height, the initial kinetic energy has become additional gravitational energy. The starting gravitational energy is already present at 0.40 m0.40\ \mathrm{m}.
    12mvi2+mghi=mghf\frac{1}{2}mv_i^2+mgh_i=mgh_f
  3. Find the final height
    Substitute the values and isolate the final height. The mass cancels, and the result is rounded to two significant figures.
    hf=0.40 m+(5.0 m/s)22(9.8 N/kg)=1.7 mh_f=0.40\ \mathrm{m}+\frac{(5.0\ \mathrm{m/s})^2}{2(9.8\ \mathrm{N/kg})}=1.7\ \mathrm{m}
Answer: The object reaches an estimated height of 1.7 m1.7\ \mathrm{m} above the floor.
Check: The added height is about 1.3 m1.3\ \mathrm{m}, which is consistent with the initial upward speed. The units in the added-height term reduce to metres. The height is positive and greater than the starting height, as expected.

Common mistakes and how to avoid them

Treating downward velocity as negative kinetic energy.
Correction: Velocity has direction, but kinetic energy uses speed squared and is not negative.
Changing the height reference between the initial and final calculations.
Correction: Use one reference level for both heights. Changes in gravitational energy then compare correctly.
Assuming measured energy changes must match perfectly.
Correction: Compare the values and consider measurement limits and transfers to the surroundings. Do not label a prediction as measured evidence.

Lesson summary

Check your understanding

Question 1

An object moves downward faster while falling. Which statement best describes its energy transformation in the simple model?
  1. Gravitational energy increases and kinetic energy decreases.
  2. Gravitational energy decreases and kinetic energy increases.
  3. Both gravitational and kinetic energy decrease.
  4. Kinetic energy becomes negative because the object moves downward.
Show answer and explanation
Gravitational energy decreases and kinetic energy increases.
As height decreases, gravitational energy decreases. As speed increases, kinetic energy increases. Kinetic energy is not negative for downward motion.

Question 2

A moving object's speed doubles while its mass stays the same. What happens to its kinetic energy?
  1. It doubles.
  2. It becomes four times as large.
  3. It is cut in half.
  4. It does not change.
Show answer and explanation
It becomes four times as large.
Kinetic energy depends on speed squared. Doubling speed multiplies the energy by four.

Question 3

In a proposed drop investigation, which item is measured evidence?
  1. A predicted speed calculated from height alone.
  2. A simulated energy value.
  3. The speed recorded by a photogate during an actual trial.
  4. The assumption that all energy transfers are negligible.
Show answer and explanation
The speed recorded by a photogate during an actual trial.
A photogate reading from an actual trial is measured evidence. Predictions, simulation outputs, and assumptions are not measurements from that trial.

Key terms

Gravitational energy
Energy an object has because of its height in a gravitational field.
Kinetic energy
Energy an object has because it is moving.
Scalar
A quantity with magnitude but no direction.
Vector
A quantity with both magnitude and direction.
Reference level
The chosen height where the height value is set to zero.
SI unit
A standard measurement unit used in science, such as the metre, kilogram, or joule.

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Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 11 Physics (SPH3U), expectation D2.4. It is a study resource, not an official curriculum publication.

Before publication, the draft is checked for structure, mathematical or chemical notation, calculations, course boundaries, and readability, and then requires administrator approval. Errors can still occur, so corrections are welcomed.

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