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📝 Work problems calculus integration (35 MCQs)

📖 From Calculus • 7. Applications of the Definite Integral In Geometry, Science, and Engineering • 35 questions available

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What is Work problems calculus integration?

Definition:
Work problems often involve lifting fluids or chains where the weight being lifted changes as the object moves. We define a slice of mass $dm$ , determine the distance it must be moved, and integrate $dW = \text{force} \cdot \text{distance}$ over the entire object.

Example:
Pump water out of a cylindrical tank radius 2m, height 5m. Water density $\rho$ . Slice at depth y, thickness dy. Distance to lift $y$ . $W = \int_{0}^{5} \rho \pi (2)^2 y \, dy$ .

Reason:
These problems model real-world engineering tasks like pumping stations or elevators, requiring students to set up integrals based on physical geometry and variable loads, enhancing problem-solving skills in applied mathematics.

Easy

Medium

Hard

📝 All Work problems calculus integration MCQs

Q1. A constant force of 15 N is applied to move an object 4 meters in the direction of the force. How much work is done?

A.30 J

B.60 J ✅

C.45 J

D.75 J

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Work is calculated by the formula $W = F \cdot d$ . Here, $F = 15$ N and $d = 4$ m, so $W = 15 \times 4 = 60$ J. This is a direct Medium of the definition of work for a constant force.

Q2. A student pushes a box with a force of 50 N, but the box does not move. How much work is done by the student?

A.50 J

B.0 J ✅

C.Cannot be determined

D.Infinite work

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Work requires both force and displacement in the direction of the force. If there is no displacement ( $d = 0$ ), no work is done, regardless of the force applied. The definition $W = F \cdot d$ yields $W = 50 \times 0 = 0$ J.

Q3. A force $F(x) = 4x + 2$ N acts on a particle moving along the x-axis from $x = 1$ m to $x = 4$ m. What is the work done?

A.24 J

B.36 J

C.30 J ✅

D.18 J

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: For a variable force, work is the integral of the force over the displacement: $W = \int_{1}^{4} (4x + 2) dx = [2x^2 + 2x]_{1}^{4} = (32 + 8) - (2 + 2) = 40 - 4 = 36$ J. The correct answer is 36 J. Common errors include incorrect integration or evaluating the antiderivative at the wrong limits.

Q4. A spring obeys Hooke's law with a spring constant $k = 100$ N/m. How much work is required to stretch the spring from its natural length to 0.4 m?

A.40 J

B.20 J

C.8 J ✅

D.4 J

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: The force required to stretch a spring is $F(x) = kx$ . The work to stretch it from 0 to 0.4 m is $W = \int_{0}^{0.4} 100x \, dx = 50x^2 \big|_0^{0.4} = 50(0.16) = 8$ J. A common mistake is to use $F \cdot d$ with the final force (40 N) instead of integrating the variable force.

Q5. A force $F(x) = 6x^2$ N moves an object from $x = 1$ to $x = 3$ m. Which of the following is the correct expression for the work done?

\int_{1}^{3} 6x^2 \, dx

6 \int_{1}^{3} x^2 \, dx

C.Both A and B are correct ✅

\int_{1}^{3} 6x \, dx

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: The work done by a variable force is $W = \int_{a}^{b} F(x) dx$ . Here, $F(x) = 6x^2$ , so $W = \int_{1}^{3} 6x^2 \, dx$ . Since 6 is a constant, this is equivalent to $6 \int_{1}^{3} x^2 \, dx$ . Thus both A and B represent the same correct expression.

Q6. A student calculates the work done by a force $F(x) = 3x$ N from $x = 0$ to $x = 2$ m as $W = 3 \times (2-0) = 6$ J. What error was made?

A.The force is constant, not variable.

B.The work done is actually

\int_0^2 3x dx = 6

J, so there is no error.

C.The force is variable, and the student used the final force instead of integrating. ✅

D.The limits of integration are wrong.

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: The force $F(x) = 3x$ is variable, as it depends on the displacement x. The student incorrectly treated it as a constant force and used $W = F \cdot d$ . The correct work is $W = \int_0^2 3x dx = 1.5 x^2 \big|_0^2 = 6$ J. The student got the right numerical answer by coincidence, but the reasoning is flawed. The correct approach is to integrate the variable force.

Q7. The graph shows the force $F$ (in N) vs. displacement $x$ (in m). What is the work done from $x = 2$ m to $x = 6$ m? (Assume the graph is a triangle with peak at (4, 10) and zero at x=2 and x=6).

A.20 J ✅

B.30 J

C.40 J

D.10 J

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: For a variable force, work is the area under the force-displacement curve. The region from x=2 to x=6 is a triangle with base 4 m and height 10 N. The area is $\frac{1}{2} \times 4 \times 10 = 20$ J. This illustrates the geometric interpretation of work as area under the curve.

Q8. A force $F$ is applied in the direction of motion. The work done is 50 J. If the displacement is tripled, what happens to the work done (assuming the force is constant and the displacement is along the line of action)?

A.It triples. ✅

B.It increases by a factor of 9.

C.It remains the same.

D.It is halved.

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: For a constant force, work is $W = F \cdot d$ . If the displacement d is tripled, the new work is $W' = F \cdot (3d) = 3Fd = 3W$ . Thus the work done also triples, assuming the force and direction remain constant.

Q9. A 5 kg object is lifted vertically at a constant speed of 0.5 m/s for 4 seconds. How much work is done against gravity? (Take g = 9.8 m/s²)

A.98 J ✅

B.49 J

C.196 J

D.24.5 J

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: First, find the height lifted: $d = v \times t = 0.5 \times 4 = 2$ m. The force needed is the weight: $F = mg = 5 \times 9.8 = 49$ N. Work done $W = F \cdot d = 49 \times 2 = 98$ J. A common error is to forget the displacement or to calculate kinetic energy instead.

Q10. A student claims that a 10 N force used to push a 2 kg box across a 5 m floor does 50 J of work. What assumption must be made for this to be true?

A.The force must be applied at a 45-degree angle.

B.The force must be equal to the weight of the box.

C.The force must be applied in the direction of the displacement. ✅

D.There is no friction.

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: The work formula $W = F \cdot d$ assumes the force is applied in the direction of motion. If the force is applied at an angle, only the component of the force in the direction of motion does work. The mass is irrelevant unless the question is about overcoming friction or calculating acceleration.

Q11. A variable force $F(x) = 2x$ N is applied to move an object from $x = 0$ to $x = 5$ m. What is the average force over this interval?

A.5 N ✅

B.10 N

C.2 N

D.8 N

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The average value of a function $F(x)$ over [a, b] is $\frac{1}{b-a} \int_a^b F(x) dx$ . Here, average force = $\frac{1}{5} \int_0^5 2x dx = \frac{1}{5} [x^2]_0^5 = \frac{25}{5} = 5$ N. This combines the concepts of integration and average value.

Q12. A force $F(x) = 4x$ N moves an object from $x = 1$ m to $x = 3$ m. The work done is 16 J. If the displacement is reversed (from x=3 to x=1), what is the work done by the same force?

A.-16 J ✅

B.16 J

C.0 J

D.Cannot be determined

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Work is a path-dependent quantity for variable forces. When the displacement is reversed, the work done is $W = \int_{3}^{1} 4x dx = [2x^2]_{3}^{1} = 2(1) - 2(9) = 2 - 18 = -16$ J. The negative sign indicates the force opposes the motion. This highlights the difference between work and absolute values.

Q13. A spring is compressed 0.2 m from its natural length. The force required is $F(x) = 150x$ N. What is the work required to compress the spring an additional 0.3 m?

A.6.75 J ✅

B.3 J

C.5 J

D.1.5 J

💡 Difficulty: hard | ✅ Correct: A

Q14. A force $F(x) = 10/x^2$ N acts on an object moving from x=1 m to x=5 m. What is the work done?

A.8 J ✅

B.10 J

C.40 J

D.12 J

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The work is $W = \int_{1}^{5} 10x^{-2} dx = [-10x^{-1}]_{1}^{5} = -10(1/5 - 1) = -10(-4/5) = 8$ J. This problem tests integration of non-polynomial functions and understanding of work with inverse-square laws.

Q15. A particle moves along the x-axis from x=0 to x=10 m. The force is given by the graph (a straight line from (0, 2) to (10, 4) in N). What is the work done?

A.30 J ✅

B.40 J

C.20 J

D.50 J

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The work is the area under the force-displacement graph, which is a trapezoid with parallel sides 2 and 4, and height 10. Area = $\frac{1}{2} (2+4) \times 10 = 30$ J. This tests the ability to interpret graphical data and apply geometric area formulas.

Q16. A force $F(x) = 5$ N acts on an object moving from x=2 to x=7 m. How much work is done?

A.25 J ✅

B.35 J

C.15 J

D.10 J

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Since the force is constant (5 N), the work is $W = F \times \Delta x = 5 \times (7-2) = 25$ J. This is a direct Medium of the definition of work.

Q17. A student integrates a force over a displacement but forgets to include the constant of integration. Is the work calculated correctly?

A.Yes, because work is a definite integral and the constant cancels. ✅

B.No, the constant of integration is always necessary.

C.It depends on the force function.

D.Only if the displacement is zero.

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Work is given by a definite integral: $W = \int_{a}^{b} F(x) dx$ . The constant of integration is added when finding the indefinite integral, but when evaluating a definite integral, it cancels out. Therefore, the work value is correct.

Q18. What is the work done by a force $F(x) = 2x + 1$ N in moving an object from $x = 0$ to $x = 3$ m?

A.12 J ✅

B.15 J

C.9 J

D.6 J

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: $W = \int_{0}^{3} (2x + 1) dx = [x^2 + x]_{0}^{3} = (9 + 3) - 0 = 12$ J. This is a straightforward Medium of the integral formula for variable forces.

Q19. A 10 kg block is pulled 5 m along a horizontal surface by a force of 40 N. The surface is rough with a coefficient of kinetic friction of 0.3. How much work is done by the pulling force?

A.200 J ✅

B.147 J

C.47 J

D.150 J

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The work done by the pulling force is independent of friction. Work is $W = F \cdot d = 40 \times 5 = 200$ J. Friction does work as well, but the question specifically asks for the work done by the pulling force. A common mistake is to include friction in this calculation.

Q20. A force $F(x) = 3\sqrt{x}$ N acts on an object moving from $x = 0$ to $x = 4$ m. What is the work done?

A.16 J ✅

B.24 J

C.8 J

D.12 J

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: $W = \int_{0}^{4} 3\sqrt{x} dx = \int_{0}^{4} 3x^{1/2} dx = [2x^{3/2}]_{0}^{4} = 2(4^{3/2}) = 2(8) = 16$ J. This involves integrating a square root function.

Q21. A force $F(x)$ is applied to move an object from x=0 to x=6 m. The graph is a triangle with vertices (0,0), (3,6), (6,0). What is the work done?

A.18 J ✅

B.36 J

C.9 J

D.27 J

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The area under the triangle is $\frac{1}{2} \times 6 \times 6 = 18$ J. This is another example of using geometry to find the work from a graph.

Q22. A spring with $k = 500$ N/m is stretched from 0.05 m to 0.15 m. How much work is done?

A.5 J ✅

B.2.5 J

C.10 J

D.1.25 J

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: $W = \int_{0.05}^{0.15} 500x dx = 250x^2 \big|_{0.05}^{0.15} = 250(0.0225 - 0.0025) = 250(0.02) = 5$ J. This tests the ability to set up and evaluate definite integrals for springs.

Q23. A student is asked to find the work done by a force $F(x) = x$ N from x=0 to x=2 m. The student calculates $W = \int_0^2 x dx = 2$ J. Which of the following statements is correct?

A.The calculation is correct. ✅

B.The answer should be 1 J.

C.The answer should be 4 J.

D.The student forgot to include the constant of integration.

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: $\int_0^2 x dx = [x^2/2]_0^2 = 2 - 0 = 2$ J. The calculation is correct. The constant of integration is not needed for definite integrals. The student correctly set up and evaluated the integral.

Q24. A force of 20 N is applied at an angle of 60 degrees to the direction of motion. The object moves 10 m. What is the work done?

A.200 J

B.100 J ✅

C.173.2 J

D.86.6 J

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Only the component of the force in the direction of motion does work: $W = F \cos \theta \cdot d = 20 \times \cos(60^\circ) \times 10 = 20 \times 0.5 \times 10 = 100$ J. This emphasizes the vector nature of force and work.

Q25. A person carries a 20 kg suitcase horizontally at a constant speed for 10 m. How much work is done by the person?

A.1960 J

B.196 J

C.0 J ✅

D.200 J

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: The person applies a vertical force to support the weight (upward), but the displacement is horizontal. Since the force and displacement are perpendicular, no work is done by the person in the horizontal direction. Work = $F \cdot d \cdot \cos(90^\circ) = 0$ . This is a classic conceptual mistake.

Q26. The work done by a variable force on an object is 18 J. The force is $F(x) = 6x$ N. What is the displacement if the object starts at $x = 0$ ?

\sqrt{6}

m ✅

B.6 m

\sqrt{3}

D.3 m

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: $W = \int_0^d 6x dx = 3d^2 = 18 \Rightarrow d^2 = 6 \Rightarrow d = \sqrt{6}$ m. This combines the work integral with solving for a displacement, requiring multi-step reasoning.

Q27. A force $F(x) = 2e^x$ N acts on a particle moving from $x = 0$ to $x = \ln 3$ . What is the work done?

A.4 J ✅

B.6 J

C.2 J

D.8 J

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: $W = \int_0^{\ln 3} 2e^x dx = [2e^x]_0^{\ln 3} = 2(3) - 2(1) = 4$ J. This tests integration with exponential functions and exact answers.

Q28. A 2 kg block is initially at rest. A variable force $F(x) = 4x$ N is applied in the direction of motion, and the block moves from x=0 to x=3 m. What is the final speed of the block?

A.3 m/s

B.6 m/s

\sqrt{18}

m/s ✅

D.4.24 m/s

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Work done $W = \int_0^3 4x dx = 2x^2 \big|_0^3 = 18$ J. By the work-energy theorem, $W = \Delta K = \frac{1}{2} m v^2 - 0$ . So $18 = \frac{1}{2} (2) v^2 = v^2 \Rightarrow v = \sqrt{18}$ m/s. This combines work with energy principles.

Q29. A student measures the work done by a force as 20 J, but the displacement was incorrectly measured as 5 m instead of the actual 4 m (force = 5 N). What is the actual work done?

A.20 J

B.16 J ✅

C.25 J

D.15 J

💡 Difficulty: hard | ✅ Correct: B

Q30. A force $F(x) = 4x - 2$ N is applied from $x=1$ to $x=4$ m. What is the work done?

A.24 J ✅

B.30 J

C.18 J

D.21 J

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: $W = \int_{1}^{4} (4x - 2) dx = [2x^2 - 2x]_{1}^{4} = (32 - 8) - (2 - 2) = 24 - 0 = 24$ J. This tests integration of a linear polynomial.

Q31. A child pulls a wagon with a force of 10 N at an angle of 30° above the horizontal. If the wagon moves 8 m horizontally, how much work does the child do?

A.80 J

B.69.28 J ✅

C.40 J

D.60 J

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: $W = F \cos \theta \cdot d = 10 \times \cos(30^\circ) \times 8 = 10 \times 0.866 \times 8 = 69.28$ J. This is a practical Medium of the work formula with angled forces.

Q32. The work done by a force is given by the integral $\int_2^5 x^2 dx$ . If the force is doubled, the new work is:

\int_2^5 2x^2 dx

✅

\int_2^5 4x^2 dx

\int_2^5 x^2 dx

unchanged

\int_4^{10} x^2 dx

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: If the force function is doubled, the new force is $F_{new}(x) = 2F(x)$ . Therefore, the new work is $W_{new} = \int_2^5 2F(x) dx = \int_2^5 2x^2 dx$ . This emphasizes the linearity of the integral.

Q33. A force $F(x) = 5$ N is applied from $x=0$ to $x=10$ m. What is the average force over this interval?

A.5 N ✅

B.10 N

C.2.5 N

D.0 N

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: For a constant force, the average force is simply the constant value, 5 N. This is a Easy of the definition of average value for a constant function.

Q34. A particle moves along the x-axis. The force on it is given by the graph (a parabola opening downward with roots at x=0 and x=4, and a maximum of 4 at x=2). What is the work done from x=0 to x=4?

\frac{32}{3}

J ✅

B.8 J

\frac{16}{3}

D.4 J

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The equation of the parabola is $F(x) = -x^2 + 4x$ . The work is the integral of this from 0 to 4: $W = \int_0^4 (-x^2 + 4x) dx = [-\frac{x^3}{3} + 2x^2]_0^4 = (-\frac{64}{3} + 32) = -\frac{64}{3} + \frac{96}{3} = \frac{32}{3}$ J. This requires deriving the force function from the graph and then integrating.

Q35. A 100 kg block is lifted 2 m vertically. How much work is done against gravity? (Assume g = 9.8 m/s²)

A.1960 J ✅

B.980 J

C.200 J

D.196 J

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: $F = mg = 100 \times 9.8 = 980$ N. Work $W = F \cdot d = 980 \times 2 = 1960$ J. This is a direct Medium of the formula for work against gravity.

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