📝 Conservation of energy vector calculus (16 MCQs)

📖 From Calculus • 16. Topics in vector Calculus • 16 questions available

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What is Conservation of energy vector calculus?

Conservation of energy vector calculus:
In a conservative field, the total mechanical energy $E = K + U$ is conserved, where $U = -f$ is potential energy, and work $W = \Delta K = -\Delta U$ .

Example:
For gravity $\mathbf{F} = \langle 0, -mg \rangle$ , $f = -mgy$ , so $U = mgy$ , and $K + mgy$ remains constant.

Reason:
Vector calculus formalizes energy conservation, linking mathematics to fundamental physical laws.

Easy

Medium

Hard

📝 All Conservation of energy vector calculus MCQs

Q1. A particle moves in a region where the net force is conservative. Its kinetic energy changes from $18\text{ J}$ to $7\text{ J}$ . Which conclusion about the change in potential energy is necessarily correct?

A.It decreases by

11\text{ J}

B.It increases by

11\text{ J}

✅

C.It increases by

25\text{ J}

D.It remains unchanged

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: For a system influenced only by conservative forces, the total mechanical energy remains constant. Therefore, any decrease in kinetic energy must appear as an equal increase in potential energy. Here the kinetic energy decreases by $18-7=11\text{ J}$ , so the potential energy increases by $11\text{ J}$ .

Q2. A student claims that conservation of energy means the kinetic energy of a particle must remain constant whenever the force field is conservative. Which evaluation is most accurate?

A.Correct, because conservative forces cannot do work

B.Correct, because potential energy is always zero

C.Incorrect, because kinetic and potential energy can exchange while their sum remains constant ✅

D.Incorrect, because conservative forces always increase kinetic energy

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: Conservation of mechanical energy does not require kinetic energy to remain constant. A conservative force can convert potential energy into kinetic energy or kinetic energy into potential energy. The essential condition is that the sum $K+U$ remains constant when no nonconservative work changes the mechanical energy.

Q3. A particle is released from rest at a point where its potential energy is $42\text{ J}$ . Later, its potential energy is $15\text{ J}$ , and no nonconservative work acts. What is its kinetic energy at the later point?

A.15 J

B.27 J ✅

C.42 J

D.57 J

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Initially the particle has total mechanical energy $E=K+U=0+42=42\text{ J}$ . At the later point, conservation gives $42=K+15$ . Therefore $K=27\text{ J}$ . The key reasoning is to track the total energy rather than treating potential and kinetic energies separately as conserved.

Q4. Two different paths connect the same initial and final positions in a conservative force field. Along path A the particle gains $24\text{ J}$ of kinetic energy, while along path B it loses $6\text{ J}$ of kinetic energy. What can be concluded?

A.Both paths are possible only if the force is nonconservative

B.The potential-energy change must be different for the two paths

C.The stated kinetic-energy changes cannot both correspond to the same initial and final states under only a conservative force ✅

D.The total mechanical energy is different for each path

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: For a conservative force field, potential energy depends only on position, so the change in potential energy between fixed endpoints is unique. With no nonconservative work, the kinetic-energy change must therefore also be unique. Changes of $+24\text{ J}$ and $-6\text{ J}$ cannot both describe the same endpoints under those assumptions.

Q5. A cart moves through a track with a conservative force and reaches a point where its speed is smaller than at the starting point. A student concludes that energy has been destroyed. Which explanation best corrects the reasoning?

A.The missing kinetic energy has necessarily become potential energy ✅

B.The force must have stopped acting

C.The cart has violated conservation of energy

D.A conservative force cannot change speed

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: A conservative force can transfer energy between kinetic and potential forms without changing total mechanical energy. If the cart slows down, its kinetic energy decreases, but the potential energy can increase by the same amount. Thus a decrease in speed does not imply energy destruction.

Q6. A particle has total mechanical energy $60\text{ J}$ . At position $x_1$ , its potential energy is $20\text{ J}$ . At position $x_2$ , the potential energy is $45\text{ J}$ . If only conservative forces act, how do the kinetic energies compare?

K_2

25\text{ J}

greater than

K_1

K_2

25\text{ J}

less than

K_1

✅

K_2=K_1

D.Both kinetic energies must be zero

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: At $x_1$ , $K_1=60-20=40\text{ J}$ . At $x_2$ , $K_2=60-45=15\text{ J}$ . Therefore $K_2$ is $25\text{ J}$ less than $K_1$ . The increase in potential energy exactly matches the decrease in kinetic energy because total mechanical energy is fixed.

Q7. A roller-coaster car starts with mechanical energy $500\text{ J}$ . Friction does $80\text{ J}$ of negative work before the car reaches a later position where its potential energy is $270\text{ J}$ . What is its kinetic energy there?

A.150 J ✅

B.230 J

C.310 J

D.420 J

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Negative work by friction removes mechanical energy from the car. Thus the later mechanical energy is $500-80=420\text{ J}$ . Since $E=K+U$ , the kinetic energy is $K=420-270=150\text{ J}$ . The important distinction is between total energy conservation and conservation of mechanical energy.

Q8. A student uses $K+U=\text{constant}$ for a motion in which an external agent continuously supplies energy to the system. The calculated final speed is much too small. What is the most likely modeling error?

A.The student assumed kinetic energy can never change

B.The student ignored energy transferred into the system by the external agent ✅

C.The student treated potential energy as a vector

D.The student assumed position affects potential energy

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The relation $K+U=\text{constant}$ applies when no nonconservative external work changes the mechanical energy of the modeled system. If an external agent supplies energy, the mechanical-energy balance must include that energy transfer. Omitting the positive external work makes the predicted final kinetic energy, and therefore speed, too small.

Q9. A graph of potential energy $U(x)$ rises from $10\text{ J}$ at $x=1$ to $35\text{ J}$ at $x=4$ . A particle has constant total mechanical energy $40\text{ J}$ . Which statement best describes its kinetic energy over this interval?

A.It increases from

30\text{ J}

5\text{ J}

B.It decreases from

30\text{ J}

5\text{ J}

✅

C.It remains

40\text{ J}

throughout

D.It increases from

5\text{ J}

30\text{ J}

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Mechanical energy satisfies $E=K+U$ , so $K=E-U$ . At $x=1$ , $K=40-10=30\text{ J}$ . At $x=4$ , $K=40-35=5\text{ J}$ . Thus the rising potential-energy graph corresponds to decreasing kinetic energy, assuming the particle remains dynamically allowed.

Q10. Two students solve the same conservative-force problem. Student A calculates the final speed using forces and acceleration, while Student B uses $K+U=\text{constant}$ . They obtain different answers. Which approach provides the strongest diagnostic strategy?

A.Always trust the force method

B.Always trust the energy method

C.Check whether both methods use the same initial conditions, potential-energy reference, and assumptions about nonconservative work ✅

D.Average the two calculated speeds

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Both methods can be valid, but disagreement usually indicates inconsistent modeling. The students should compare initial and final conditions, the definition of potential energy, and whether any nonconservative work is present. Changing the zero level of potential energy is harmless if done consistently, but omitting energy transfers is not.

Q11. A particle moves in a potential-energy landscape. At one point $U=12\text{ J}$ , and at another point $U=28\text{ J}$ . Its total mechanical energy is $25\text{ J}$ . What does the energy model predict about reaching the second point?

A.It reaches it with

3\text{ J}

of kinetic energy

B.It reaches it with

53\text{ J}

of kinetic energy

C.It cannot reach it under the stated energy conditions ✅

D.It reaches it only if potential energy becomes negative

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: At the second point, conservation would require $K=E-U=25-28=-3\text{ J}$ . Negative kinetic energy is not physically possible in this classical model. Therefore the particle cannot reach that position with total mechanical energy $25\text{ J}$ . This identifies an energetically forbidden region.

Q12. A particle is moving through a conservative potential with total energy $E$ . A graph shows $U(x)$ touching $E$ at $x=a$ and $x=b$ , while $U(x)<E$ between them. Which interpretation is most appropriate?

A.The particle has maximum speed at

a

and

b

B.The particle's kinetic energy is zero at

a

and

b

, and motion between them is energetically allowed ✅

C.The particle cannot move between

a

and

b

because potential energy changes

D.The particle has constant kinetic energy throughout the interval

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Because $K=E-U$ , points where $U=E$ have $K=0$ . Thus $x=a$ and $x=b$ are turning points under the idealized one-dimensional model. Between them, $U<E$ , so $K>0$ and the motion is energetically allowed.

Q13. A system contains a particle moving under a conservative force and a spring. The particle loses $12\text{ J}$ of kinetic energy while the spring gains $7\text{ J}$ of elastic potential energy. If no other energy transfer occurs, what must happen to the remaining $5\text{ J}$ ?

A.It must disappear because energy conservation applies only to kinetic energy

B.It must become another form of energy or indicate that the stated measurements are incomplete ✅

C.It must increase the particle's mass

D.It must increase the spring's stiffness automatically

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The total energy balance requires all changes to be accounted for. A $12\text{ J}$ decrease in kinetic energy accompanied by only $7\text{ J}$ of potential-energy increase leaves $5\text{ J}$ unaccounted for. That amount must appear in another energy form, such as thermal energy, or signal incomplete or inaccurate modeling.

Q14. A particle moves between two points in a conservative field. At the first point its speed is $4\text{ m/s}$ , and at the second it is $10\text{ m/s}$ . If its mass is $2\text{ kg}$ , what is the corresponding change in potential energy?

A.84 J increase

B.84 J decrease ✅

C.42 J increase

D.28 J decrease

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The kinetic-energy change is $\Delta K=\frac12m(v_2^2-v_1^2)=\frac12(2)(100-16)=84\text{ J}$ . With only conservative forces, $\Delta K+\Delta U=0$ , so $\Delta U=-84\text{ J}$ . Thus potential energy decreases by $84\text{ J}$ , supplying the increase in kinetic energy.

Q15. A system is modeled by $E=K+U$ . A proposed solution states: 'The particle speeds up, so its potential energy must increase because both forms of energy increase together.' Which revision is most consistent with the model when no external work occurs?

A.Speed and potential energy must always increase together

B.An increase in kinetic energy requires an equal decrease in potential energy ✅

C.Potential energy is unrelated to kinetic energy

D.Both energies can increase only if the particle stops

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: When mechanical energy is conserved, $K+U$ remains constant. Therefore, if kinetic energy increases, potential energy must decrease by the same amount. The student's error is assuming that different forms of energy must rise together; conservation instead constrains their sum and permits continuous energy exchange.

Q16. A particle of mass $m$ moves between two points in a conservative field. Its speed changes from $v$ to $2v$ . Without knowing $m$ , $v$ , or the absolute potential energies, what can be determined about the potential-energy change?

A.It is

+\frac32mv^2

B.It is

-\frac32mv^2

✅

C.It is

-\frac12mv^2

D.It is

+2mv^2

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The initial kinetic energy is $K_i=\frac12mv^2$ , while the final kinetic energy is $K_f=\frac12m(2v)^2=2mv^2$ . Therefore $\Delta K=2mv^2-\frac12mv^2=\frac32mv^2$ . Conservation of mechanical energy requires $\Delta U=-\Delta K=-\frac32mv^2$ .

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