Energy Practice Problems

  1. The Gibbs free energy equation is directly related to which of the following?
  1. Work and time
  2. Heat and work
  3. Enthalpy and entropy
  4. Power and enthalpy
Show Answer
The correct answer is C!

Gibbs free energy ties together enthalpy and entropy:

\(G=H-TS\)

For changes, \(\Delta G=\Delta H – T\Delta S\). At constant \(T,P\), a negative \(\Delta G\) predicts spontaneity.

 

  1. Which expression gives the change in Gibbs free energy for a process at constant temperature?
  1. \(G=H-TS\)
  2. \(\Delta G=\Delta H-T\Delta S\)
  3. \(\Delta G=\Delta H-\frac{T}{S}\)
  4. \(G=H-\frac{T}{S}\)
Show Answer
The correct answer is B!

The definition of the state function is \(G=H-TS\). The process form is \(\Delta G=\Delta H-T\Delta S\).

 

  1. Which of the following is the correct expression of Bernoulli’s principle?
  1. As KE increases, another form of energy must decrease.
  2. As KE increases, other forms of energy also increase.
  3. KE remains relatively stable in a closed loop.
  4. KE has a limited effect upon wavelength.
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The correct answer is A!

Bernoulli’s principle says that along a streamline, the sum \(P+\tfrac12\rho v^2+\rho gh\) is constant. If the kinetic term \(\tfrac12\rho v^2\) goes up (faster flow), at least one of the other forms must go down. So “as KE increases, another form of energy must decrease” matches the trade-off.

 

  1. Which of the following is the correct expression of the work–energy theorem?
  1. \(W=\Delta KE\)
  2. \(W=K-\Delta E\)
  3. \(W=K\times\Delta E\)
  4. \(W=\frac{\Delta K}{\Delta E}\)
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The correct answer is A!

The work–energy theorem states that the net work done on an object equals the change in its kinetic energy. The other expressions mis-combine quantities and are not physical laws.

 

  1. A spring has a spring constant of 120 N per meter. How much potential energy is stored in the spring as it is stretched 0.2 meters?
  1. 1.2 J
  2. 2.4 J
  3. 3.1 J
  4. 7.4 J
Show Answer
The correct answer is B!

Elastic potential energy in a spring:

\(U=\dfrac12 kx^2=\dfrac12(120)(0.20)^2=\dfrac12(120)(0.04)\) \(=60\times0.04=2.4\ \text{J}\)

 

  1. A distance of \(1 \times 10^3 \text{ m}\) separates the charge at the bottom of the cloud and the ground. The electric field intensity between the bottom of the cloud and the ground is \(2 \times 10^4\) N per coulomb. What is the potential difference between the bottom of the cloud and the ground?
  1. \(1.4 \times 10^4\text{ V}\)
  2. \(2.5 \times 10^3\text{ V}\)
  3. \(2.8 \times 10^6\text{ V}\)
  4. \(2 \times 10^7\text{ V}\)
Show Answer
The correct answer is D!

For a uniform field, potential difference is \(\Delta V=Ed\).

\(E=2\times10^4\ \text{N/C}\)

\(d=1\times10^3\ \text{m}\)

\(\Delta V=(2\times10^4)(1\times10^3)=2\times10^7\ \text{V}\)

 

  1. When Adam drinks cold water, his body warms the water until thermal equilibrium is reached. If he drinks six glasses (2.5 kilograms) of water at 0°C in a day, approximately how much energy must his body expend to raise the temperature of this water to his body’s temperature of 37°C?
  1. 210 kJ
  2. 305 kJ
  3. 390 kJ
  4. 414 kJ
Show Answer
The correct answer is C!

Heating water (no phase change):

\(Q=mc\Delta T\)

\(m=2.5\ \text{kg}\)

\(c\approx4.186\times10^3\ \text{J/(kg·°C)}\)

\(\Delta T=37^\circ\text{C}\)

\(Q\approx2.5(4.186\times10^3)(37)\) \(\approx3.87\times10^5\ \text{J}=387\ \text{kJ}\)

\(\approx 390 \text{ kJ}\)

 

  1. An electron is located between a pair of oppositely charged parallel plates. As the electron approaches the positively charged plate, what happens to the kinetic energy of the electron?
  1. It increases.
  2. It decreases.
  3. It remains the same.
  4. It transforms into potential energy.
Show Answer
The correct answer is A!

As a negative electron moves toward the positive plate, its electric potential energy \(U=qV\) decreases (since \(q\lt0\) and \(V\) increases).

Thus, by energy conservation, its kinetic energy increases.

 

  1. If the speed of a moving object is doubled, which quantity also associated with the object must double?
  1. Momentum
  2. KE
  3. Gravitational potential energy
  4. Acceleration
Show Answer
The correct answer is A!

Momentum is directly proportional to speed. Doubling \(v\) doubles \(p\).

Kinetic energy (\(KE=\tfrac12 mv^2\)) would quadruple, not double. Gravitational PE doesn’t depend on speed, and acceleration isn’t determined by speed alone.

 

  1. A 45 kg bicyclist climbs a hill at a constant speed of 2.5 meters per second by applying an average force of 85 N. Approximately how much power does the bicyclist develop?
  1. 115 W
  2. 210 W
  3. 250 W
  4. 320 W
Show Answer
The correct answer is B!

Power for constant-speed motion under force \(F\) is:

\(P=Fv=(85\ \text{N})(2.5\ \text{m/s})=212.5\ \text{W}\) \(\approx210\ \text{W}\)

 

  1. A person kicks in a 4 kg door with 48 N of force, causing the door to accelerate at 12 m/s2. What is the magnitude of the force exerted by the door on the person?
  1. 24 N
  2. 35 N
  3. 42 N
  4. 48 N
Show Answer
The correct answer is D!

The door’s acceleration confirms the applied force. By Newton’s third law, the door exerts an equal and opposite force of 48 N on the person.

 

  1. A 60 kg student running at 3 meters per second has a kinetic energy of…
  1. 111 J
  2. 151 J
  3. 260 J
  4. 270 J
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The correct answer is D!

\(KE=\dfrac12 mv^2=\dfrac12(60)(3^2)=30\times9=270\ \text{J}\)

 

  1. How much work is done in moving 5 coulombs of charge against a potential difference of 12 volts?
  1. 30 J
  2. 60 J
  3. 400 J
  4. 500 J
Show Answer
The correct answer is B!

Electrical work is:

\(W=q\Delta V=(5\ \text{C})(12\ \text{V})=60\ \text{J}\)

 

  1. Compared to insulators, metals are better conductors of electricity because metals contain more free…
  1. Protons
  2. Electrons
  3. Neutrons
  4. Positive ions
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The correct answer is B!

Metals conduct well because they have many delocalized (free) electrons in a conduction band. Protons and neutrons are bound in nuclei, and positive ions are not free charge carriers in metallic solids.

 

  1. How much time is required for an operating 100 W light bulb to dissipate 10 J of electrical energy?
  1. 1 second
  2. 0.1 seconds
  3. 0.25 seconds
  4. 200 seconds
Show Answer
The correct answer is B!

Power is energy per time:

\(P=\dfrac{E}{t}\Rightarrow t=\dfrac{E}{P}\) \(=\dfrac{10\ \text{J}}{100\ \text{W}}=0.10\ \text{s}\)