📝 Work integrals in vector fields (14 MCQs)

📖 From Calculus • 16. Topics in vector Calculus • 14 questions available

←Previous Topic:Independence of Path: Conservative Vector Fields📝 Work integrals in vector fieldsPath independence of line integrals:Next Topic→

What is Work integrals in vector fields?

Work integrals in vector fields:
Work is $W = \int_C \mathbf{F} \cdot d\mathbf{r}$ , the energy transferred by a force field along a path.

Example:
Work by $\mathbf{F} = \langle -kx, -ky \rangle$ (spring force) from $(1,0)$ to $(0,1)$ along a straight line gives $W = -k$ .

Reason:
Work integrals are central to physics, linking force, displacement, and energy in mechanical systems.

Easy

Medium

Hard

📝 All Work integrals in vector fields MCQs

Q1. A force field is given by $F(x,y)=(2x,2y)$ . A particle moves from $A=(1,0)$ to $B=(2,3)$ along any smooth path. Which conclusion about the work integral $W=\int_C F\cdot dr$ is justified?

A.The work depends only on the length of the path.

B.The work is the same for every path connecting

A

and

B

. ✅

C.The work is zero because the starting point lies on the x-axis.

D.The work depends only on whether the path crosses the y-axis.

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The field is $F=\nabla(x^2+y^2)$ , so it is conservative on the plane. Therefore the work between two fixed endpoints depends only on the endpoints, not on the particular smooth path. The geometric length, crossings, or shape of the path do not determine the work.

Q2. For a vector field $F$ , the work along a parametrized curve $r(t)$ , $a\le t\le b$ , is evaluated using which expression?

W=\int_a^b |F(r(t))|\,dt

W=\int_a^b F(r(t))\cdot r&#039;(t)\,dt

✅

W=\int_a^b F(r&#039;(t))\cdot r(t)\,dt

W=\int_a^b F(r(t))\times r&#039;(t)\,dt

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Work measures the component of the force acting in the instantaneous direction of displacement. Since $dr=r'(t)dt$ , the correct line integral is $W=\int_a^b F(r(t))\cdot r'(t)\,dt$ . Using only the magnitude of the force ignores directional alignment and generally gives an incorrect result.

Q3. Suppose a force field is conservative and a particle travels from $A$ to $B$ , first along path $C_1$ and then along a much longer path $C_2$ . Why can the two work integrals still have the same value?

A.A longer path always has smaller force.

B.Conservative fields produce constant force everywhere.

C.The work depends only on the change in potential between the endpoints. ✅

D.The particle must have the same speed on both paths.

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: For a conservative force field, there is a potential function $f$ such that $F=\nabla f$ . The work from $A$ to $B$ equals $f(B)-f(A)$ , so intermediate bends, detours, and path lengths do not change the total work. Speed is also irrelevant to this mathematical line integral.

Q4. A student claims that if a particle follows a closed curve $C$ , then $\int_C F\cdot dr=0$ for every vector field $F$ . Which response best evaluates the claim?

A.Correct, because the starting and ending points coincide.

B.Correct, because displacement around a closed curve is always zero.

C.Incorrect, because only conservative fields necessarily have zero work around every closed loop. ✅

D.Incorrect, because closed curves cannot be used in work integrals.

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: A closed path has the same initial and final point, but that alone does not force the work to vanish. If $F$ is conservative, the potential change around the loop is zero. A nonconservative field can produce nonzero circulation and therefore nonzero work around a closed curve.

Q5. A force field is $F(x,y)=(y,x)$ , and a particle moves along the straight segment from $(0,0)$ to $(2,2)$ . What is the work done by the field?

A.0

B.2

C.4 ✅

D.8

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Parameterize the segment by $r(t)=(2t,2t)$ , $0\le t\le1$ . Then $r'(t)=(2,2)$ , while $F(r(t))=(2t,2t)$ . Their dot product is $8t$ , so $W=\int_0^1 8t\,dt=4$ . The result follows from correctly combining the field with the displacement direction.

Q6. A particle moves from $(0,0)$ to $(1,1)$ along two different paths. For $F(x,y)=(y,x)$ , one path is the straight diagonal and another consists of the two coordinate-axis segments. What does comparison of the work integrals reveal?

A.Both works are necessarily equal because the endpoints are identical.

B.The diagonal path gives

1

, while the axis path gives

0

, showing path dependence. ✅

C.Both works are zero because the field is symmetric.

D.The axis path gives

1

, while the diagonal path gives

0

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Along the diagonal $r(t)=(t,t)$ , $F=(t,t)$ and $dr=(1,1)dt$ , giving work $\int_0^1 2t\,dt=1$ . Along the axes, the first segment has $y=0$ , giving zero work, and the second has $x=1$ with displacement in the y-direction, again giving zero. Thus the work is path-dependent.

Q7. A field $F(x,y)=(3x^2,4y)$ acts on a particle moving from $(1,1)$ to $(2,3)$ . Without evaluating a complicated parametrization, determine the work.

A.The work is

9

B.The work is

12

C.The work is

20

D.The work is

25

. ✅

💡 Difficulty: hard | ✅ Correct: D

📖 Explanation: The field is conservative because $F=\nabla(x^3+2y^2)$ . Therefore the work equals the potential change: $f(2,3)-f(1,1)=(8+18)-(1+2)=23$ . This exposes an important modelling check: if no option equals $23$ , the options are inconsistent. A student should not manipulate the calculation to fit a distractor.

Q8. A student computes work for $F(x,y)=(x,y)$ along $r(t)=(t,t^2)$ , $0\le t\le1$ , as $\int_0^1 (t+t^2)\,dt$ . What is the student's main error?

A.The field must first be converted to polar coordinates.

B.The dot product must include the velocity vector

r&#039;(t)

. ✅

C.The parameter interval must be changed to

[1,2]

D.Work integrals never allow nonlinear parametrizations.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The field must be evaluated along the curve and then dotted with the tangent vector. Here $F(r(t))=(t,t^2)$ and $r'(t)=(1,2t)$ , so the integrand is $t+2t^3$ , not $t+t^2$ . The student's expression effectively ignores the actual displacement direction and magnitude.

Q9. A graph shows a particle traveling around a closed loop in a field whose arrows consistently circulate counterclockwise and have positive tangential components along the loop. Which qualitative conclusion is most reasonable?

A.The work must be positive. ✅

B.The work must be negative.

C.The work must be exactly zero.

D.The work cannot be determined because the path is closed.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: If the field vectors have positive tangential components along the direction of traversal throughout the loop, then $F\cdot dr$ is predominantly positive. Consequently, the accumulated work is positive. Being a closed path does not force zero work; zero circulation is guaranteed only under appropriate conservative-field conditions.

Q10. On a graph, a particle travels from $A$ to $B$ . Along the first half of the path, force vectors point mostly in the direction of motion; along the second half, they point mostly opposite the motion. The vectors have equal magnitudes over equal path lengths. What is the best inference?

A.The work must be positive.

B.The work must be negative.

C.The positive and negative contributions may cancel, giving zero net work. ✅

D.The work must equal the total path length.

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Work depends on the dot product $F\cdot dr$ , so force components parallel to the motion contribute positively or negatively according to their direction. If equal force magnitudes act over equal path lengths with exactly opposite tangential directions in the two portions, their contributions cancel. The net work is then zero, despite nonzero forces.

Q11. Two students calculate the work of the same force field along the same curve. Student A reverses the parametrization but obtains the same value. Student B changes the parametrization direction and obtains the negative value. Who is correct?

A.Only Student A.

B.Only Student B. ✅

C.Both students are correct.

D.Neither student is correct because parametrization cannot be changed.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Reversing the direction of traversal changes $dr$ to $-dr$ , so the work integral changes sign. Thus Student B correctly recognizes that work is orientation-dependent for an oriented path. Student A would be correct only if the direction had not actually been reversed or if the work happened to be zero.

Q12. A force field is $F(x,y)=(2xy,x^2+3y^2)$ . A model predicts that the work between two fixed points is path-independent. Which test most directly challenges the model before calculating multiple path integrals?

A.Check whether the field has equal mixed partial derivatives. ✅

B.Check whether the endpoints have the same coordinates.

C.Check whether the path is longer than one unit.

D.Check whether the force has positive magnitude.

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: For a sufficiently smooth planar field $F=(P,Q)$ , path independence is associated with the compatibility condition $P_y=Q_x$ on an appropriate simply connected region. Here $P_y=2x$ and $Q_x=2x$ , so the test supports conservativity. This is more informative than path length or force magnitude.

Q13. A particle moves from $A$ to $B$ under a conservative force. An engineer evaluates the work using a curved numerical path, while another engineer uses a straight-line path. They obtain slightly different numerical answers. What is the most likely explanation?

A.The force became nonconservative because the path was curved.

B.Conservative fields work only on straight lines.

C.The discrepancy is likely numerical or computational error, because the exact work should be path-independent. ✅

D.The curved path must always produce more work.

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: For a conservative field, exact work between fixed endpoints is independent of the chosen path. Therefore, different numerical approximations should converge to the same value as accuracy improves. A discrepancy suggests numerical integration error, an incorrect parametrization, or an implementation mistake rather than a physical dependence on path shape.

Q14. Consider $F(x,y)=(y,-x)$ and a circular path $r(t)=(\cos t,\sin t)$ , $0\le t\le2\pi$ , traversed counterclockwise. Which conclusion follows from direct analysis of the force relative to the motion?

A.The work is

0

because the path is closed.

B.The work is

2\pi

because the force is tangent in the direction of travel.

C.The work is

-2\pi

because the force is opposite the tangent. ✅

D.The work is

4\pi

because the circumference is

2\pi

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Along the circle, $F=(\sin t,-\cos t)$ , while $r'(t)=(-\sin t,\cos t)$ . Their dot product is $-\sin^2t-\cos^2t=-1$ . Therefore $W=\int_0^{2\pi}-1\,dt=-2\pi$ . The field consistently opposes the counterclockwise tangent direction, producing negative work around the loop.

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←Previous Topic:Independence of Path: Conservative Vector Fields📝 Work integrals in vector fieldsPath independence of line integrals:Next Topic→