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What are the main forms of mechanical energy, and how does the conservation of energy let us solve motion problems?

Define kinetic energy, gravitational potential energy and elastic potential energy, and apply the conservation of energy to systems with and without friction, recognizing friction transfers mechanical energy to internal (thermal) energy.

A Regents Physics answer on mechanical energy and its conservation: kinetic energy, gravitational and elastic potential energy, the conservation of energy with and without friction, and how friction transfers energy to heat, using the Reference-Table equations, with worked examples.

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  1. What this topic is asking
  2. The forms of mechanical energy
  3. The conservation of energy
  4. Solving frictionless problems by energy conservation
  5. Friction and the conversion to internal energy
  6. Reference Tables note
  7. Try this

What this topic is asking

Energy and its conservation are the second great conservation law of Regents mechanics. The Physical Setting/Physics course asks you to define the forms of mechanical energy, kinetic energy, gravitational potential energy and elastic potential energy, and to apply the conservation of energy to solve motion problems, including those where friction converts mechanical energy into heat. The exam tests energy calculations and the powerful technique of equating energy at two points in the motion.

The forms of mechanical energy

Kinetic energy depends on the square of the speed, so doubling the speed quadruples the kinetic energy, a fact the Regents tests directly. Gravitational potential energy is measured relative to a chosen reference height (often the ground), and only changes in it are physically meaningful. Elastic potential energy depends on the square of the spring's displacement.

The conservation of energy

This is the most powerful problem-solving tool in the module. For a frictionless system you do not need the kinematic equations at all: just set the total mechanical energy at the start equal to the total at the end. A ball dropped from rest converts its gravitational PE entirely into KE; a pendulum swaps PE and KE back and forth; a cart on a frictionless track trades height for speed.

Solving frictionless problems by energy conservation

The method is to write

KEi+PEi=KEf+PEfKE_i + PE_i = KE_f + PE_f

and substitute. Often one term is zero: an object dropped from rest has KEi=0KE_i = 0; at the lowest point of a swing PE=0PE = 0 (taking that as the reference). For a ball falling from height hh, mgh=12mv2mgh = \tfrac{1}{2}mv^2, and the mass cancels, giving v=2ghv = \sqrt{2gh} regardless of mass, which is consistent with free fall.

Friction and the conversion to internal energy

This is a key Regents distinction: momentum is conserved in every collision, but mechanical energy is conserved only without friction (and in perfectly elastic collisions). A box sliding to a stop on a rough floor loses all its kinetic energy to heat; the energy is not destroyed, it is converted. The work done against friction, Wf=FfdW_f = F_f d, equals the mechanical energy lost.

Reference Tables note

The Reference Tables print KE=12mv2KE = \tfrac{1}{2}mv^2, PE=mghPE = mgh and PEs=12kx2PE_s = \tfrac{1}{2}kx^2 in the Mechanics section, along with the work-energy relationship W=ΔETW = \Delta E_T. The conservation of energy itself is a stated principle rather than a separate equation; you apply it by equating total energy at two points and, when friction acts, including the energy converted to heat.

Try this

Q1. A 3.03.0 kg object moves at 4.04.0 m/s. Calculate its kinetic energy. [2 points]

  • Cue. KE=12mv2=12(3.0)(4.0)2=12(3.0)(16)=24KE = \tfrac{1}{2}mv^2 = \tfrac{1}{2}(3.0)(4.0)^2 = \tfrac{1}{2}(3.0)(16) = 24 J.

Q2. State what happens to the mechanical energy "lost" when a box slides to a stop on a rough floor. [1 point]

  • Cue. It is converted to internal (thermal) energy (heat); the total energy is conserved.

Exam-style practice questions

Practice questions written in the style of NYSED exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Regents (style)2 marksPart B-2 (constructed response). A 2.02.0 kg ball is held 5.05.0 m above the ground. Calculate its gravitational potential energy relative to the ground (g=9.81g = 9.81 m/s squared). Show the equation, substitution and answer.
Show worked answer →

A 2-point constructed-response calculation using the Reference-Table equation PE=mghPE = mgh.

Equation: PE=mghPE = mgh.
Substitution: PE=(2.0)(9.81)(5.0)PE = (2.0)(9.81)(5.0).
Answer: PE=98.1PE = 98.1 J, about 9898 J.

Markers reward the equation from the tables, correct substitution with units, and the answer in joules. A common error is omitting gg or using the height as the energy directly.

Regents (style)3 marksPart C (extended response). A 0.500.50 kg ball is dropped from rest at a height of 1.81.8 m. Neglecting air resistance and taking g=9.81g = 9.81 m/s squared, (a) calculate its gravitational potential energy at the start, (b) use the conservation of energy to find its speed just before it hits the ground. Show all work.
Show worked answer →

A 3-point Part C energy-conservation problem.

(a) Initial PE (1 point): PE=mgh=(0.50)(9.81)(1.8)=8.83PE = mgh = (0.50)(9.81)(1.8) = 8.83 J.
(b) Speed (2 points): with air resistance neglected, all the PE becomes kinetic energy: KE=PE=8.83KE = PE = 8.83 J. From KE=12mv2KE = \tfrac{1}{2}mv^2, 8.83=12(0.50)v28.83 = \tfrac{1}{2}(0.50)v^2, so v2=2(8.83)0.50=35.3v^2 = \dfrac{2(8.83)}{0.50} = 35.3 and v=5.9v = 5.9 m/s.

Markers reward equating the lost PE to the gained KE and solving for the speed. A common error is forgetting the factor of one half in the kinetic energy.

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