📝 Work done by variable force integral (34 MCQs)

📖 Explanation: This is a straightforward Medium of the work integral $W = \int_{a}^{b} F(x) \, dx$. Evaluating the integral gives $\left[ x^3 - x^2 + 5x \right]_{1}^{3} = (27-9+15) - (1-1+5) = 33 - 5 = 28$ J. A common error is incorrectly calculating the antiderivative or the arithmetic. Option B is a distractor with a common miscalculation, while Option C represents a common mistake in evaluating the integral bounds.

📖 Explanation: First, find the spring constant $k = F/x = 30 / 0.05 = 600$ N/m. Work is $\int_{0.02}^{0.04} 600x \, dx = 300(0.04^2 - 0.02^2) = 300(0.0016 - 0.0004) = 300(0.0012) = 0.36$ J. The key mistake is using the total length (12 cm and 14 cm) as the displacement variable in Hooke's Law. Option C is a trap for forgetting to convert units or using the wrong limits, resulting in an answer 10 times larger.

📖 Explanation: Work is path-independent for a conservative force, but here we are explicitly calculating the integral. For $F(x) = 2x + 3$, work from 0 to 2 is $\int_0^2 (2x+3)dx = [x^2+3x]_0^2 = 4+6 = 10$ J. While the force is conservative, the work depends on the displacement limits. This question tests if students understand that the work integral's value changes with the interval. Option B (14 J) is a distractor for those who might assume a proportional relationship (14/28 = 2/4) without doing the integral.

📖 Explanation: This problem tests the concept of an improper integral. $W = \lim_{b \to \infty} \int_{1}^{b} x^{-2} dx = \lim_{b \to \infty} [-x^{-1}]_{1}^{b} = \lim_{b \to \infty} (-1/b + 1) = 1$ J. This is a finite work done by a force that approaches zero at infinity. A common misconception is to think that an infinite distance implies infinite work, but as the force becomes negligible, the total work converges to a finite value. Option B is a distractor for those who incorrectly integrate, and Option C is a trap for those ignoring the convergence of the integral.

📖 Explanation: Work is the area under the force-displacement curve. From 0 to 2 m, it's a rectangle: $2 \times 5 = 10$ J. From 2 to 4 m, it's a triangle: $0.5 \times 2 \times 5 = 5$ J. From 4 to 6 m, it's a triangle: $0.5 \times 2 \times 10 = 10$ J. Total work is $10 + 5 + 10 = 25$ J. A common mistake is to not split the graph into geometric shapes or misreading the height of the triangles. Option C (35 J) is a common distractor for those who incorrectly calculate the area of the last triangle as 20 J or add incorrectly.

📖 Explanation: The work done by $kx^2$ is $kL^3/3$, and by $kx$ is $kL^2/2$. The ratio is $(kL^3/3)/(kL^2/2) = (L/3) \times (2) = 2L/3$. This tests the ability to compare integrals and not just the form of the force. Option B is a distractor for inverting the ratio. Option C is what the ratio would be if one incorrectly canceled the $L$ terms or if $L=1$. It requires a multi-step reasoning process: setting up the integrals, computing them, and then taking the ratio.

📖 Explanation: This is a conceptual Easy question. Work is the area under the force-displacement curve, not simply the product of average force and displacement (unless force is constant). If the average force is 10 N, and total work is 50 J, then the *average* value of the integral is 10 J/m. The total displacement $L$ could be greater than 5 m if the force was sometimes below 10 N and sometimes above, but it cannot be less than 5 m if the average force is 10 N and the work is 50 J because that would imply $W < F_{avg} \times L$. However, the work is *defined* as the integral, and if the average force is 10, then by the Mean Value Theorem for Integrals, $F(c) \times L = 50$, so L = 50/F(c) = 5 m. This assumes the average value is 10. So the displacement must be 5 m. Wait, the question asks for a statement that *must* be true. If the *average* of the force function is 10 N, then the displacement is 5 m. Option B is correct. Let's re-evaluate the options. A is a distractor. B is the correct mathematical conclusion. The correct answer should be B.

📖 Explanation: Here, $F = 3(2t) = 6t$. Work is $\int F \, dx = \int F(t) v(t) \, dt$. So, $W = \int_0^3 (6t)(2t) \, dt = \int_0^3 12t^2 \, dt = 4t^3 |_0^3 = 108$ J. This question combines concepts of variable forces, kinematics, and the definition of work. A common mistake is to forget to change the variable of integration from $x$ to $t$. Option B (54 J) is a distractor for those who integrate incorrectly or forget the factor of 2. Option A is a distractor.

📖 Explanation: The work done in stretching a spring is $W = \int_0^x kx \, dx = 0.5kx^2$. Thus, work is directly proportional to the square of the extension. A common misconception, especially from experience with constant forces, is that work is proportional to force or extension. This is a key conceptual point of the work-energy theorem for springs. Option B is a trap for confusing work with force. Option C is a trap for forgetting the role of the spring constant.

📖 Explanation: If $F(x)$ is even, then the work from -5 to 0 is $\int_{-5}^0 F(x) dx$. Let $u = -x$, then $dx = -du$, and when $x = -5$, $u=5$, and when $x=0$, $u=0$. So $\int_{-5}^0 F(x) dx = -\int_{5}^{0} F(-u) du = \int_{0}^{5} F(u) du$. Thus, the total work from -5 to 5 is $2 \times 100 = 200$ J. This test Easy of definite integrals and symmetry. Option B is a common mistake, confusing odd and even functions or thinking displacement cancels out. Option C is for those who assume the work is the same over any interval of the same length.

📖 Explanation: This is a Easy question testing a fundamental definition of work in calculus-based physics. The area under a force-displacement curve is the work done. Power is the rate of doing work, represented by the product of force and velocity. Impulse is the area under a force-time graph. This question is designed to be a 15% Easy item. It is a fundamental concept that students must know.

📖 Explanation: The work is $W = \int_0^\pi \sin x \, dx = [-\cos x]_0^\pi = -\cos \pi + \cos 0 = -(-1) + 1 = 2$ J. This is a simple calculation but tests the knowledge of the anti-derivative of sine. Option B is a distractor for confusing the integral of sine with cosine or thinking the net displacement is zero. Option D is a distractor for those who incorrectly evaluate the integral.

📖 Explanation: Work is $W = \int_{1}^{2} 3x^2 dx = [x^3]_{1}^{2} = 8-1 = 7$ J. If the displacement is reversed (from $x=2$ to $x=1$), the work done by the force is $W = \int_{2}^{1} 3x^2 dx = [x^3]_{2}^{1} = 1 - 8 = -7$ J. This tests if students understand that reversing the limits of integration changes the sign of the definite integral, which corresponds to the force doing negative work or the object doing work on the agent. Option A is a distractor for forgetting the negative sign. Option C is a trap for incorrect integration.

📖 Explanation: $W = \int_0^2 (4x^3 - 3x^2 + 2x - 1) dx = [x^4 - x^3 + x^2 - x]_0^2 = (16 - 8 + 4 - 2) - 0 = 10$ J. This is a straightforward Medium of the polynomial integration. Option B is a distractor for those who make an arithmetic error. Option C is a common error for those who integrate incorrectly (e.g., forgetting the negative sign on the -x term).

📖 Explanation: The average force is $\frac{1}{3-0} \int_0^3 (5 + 2x) dx = \frac{1}{3} [5x + x^2]_0^3 = \frac{1}{3} (15 + 9) = \frac{24}{3} = 8$ N. This question tests the concept of the average value of a function. Work is also $F_{avg} \times \Delta x = 8 \times 3 = 24$ J. A common mistake is to just take the average of the endpoints $(5 + 11)/2 = 8$, which actually works for a linear function. But the calculation using the integral is the correct general method. The distractor B is for those who confuse average force with the initial force.

📖 Explanation: The new work is $\int_{a}^{b/2} 3F(x) dx = 3 \int_a^{b/2} F(x) dx$. But the displacement is halved, so the limits are now $a$ to $b/2$. Without knowing the exact form of $F(x)$, we cannot determine the work because the integral from $a$ to $b/2$ is not necessarily half of the integral from $a$ to $b$ unless $F$ is constant. For a constant force, work is $F \times d$. Tripling $F$ and halving $d$ gives $3F \times (d/2) = 1.5 Fd = 150$ J. This is a trick question. The correct answer is based on the scaling argument. However, since the question does not state $F$ is constant, the answer is 'Cannot be determined'. Let's check the options. The question says 'If the force is tripled at every point, and the displacement is halved'. If the force is a function of $x$, and we just triple the function, the work becomes W&#039; = \int_a^{b/2} 3F(x) dx = 3 \int_a^{b/2} F(x) dx. This is not necessarily $1.5 \int_a^b F(x) dx$. Therefore, the correct answer is 'Cannot be determined'. Since that option isn't available, we must select the best possible. Actually, the question is poorly phrased. If it is a specific force, we need the function. The best answer is likely '150 J' only if we assume constant force. This question is a test of the assumption of constant force in simplistic work calculations. The correct answer should be B: 150 J, but I will mark it as 'Cannot be determined'. Since the options are D, I'll choose D as the most mathematically rigorous answer.

📖 Explanation: $W = \int_1^2 x^2 dx = [x^3/3]_1^2 = 8/3 - 1/3 = 7/3$ J. A common mistake is to think the work is additive (i.e., 1/3 + 1/3 = 2/3) or to just integrate from 0 to 2 and subtract, which is correct. This question tests the fundamental theorem of calculus. Option C is a distractor for those who incorrectly assume the work per unit distance is constant. Option D is a distractor for simple arithmetic errors.

📖 Explanation: In one dimension, any force that is a function of position alone $F(x)$ is conservative. All three options are functions of position only. A conservative force in 1D can be expressed as the negative derivative of a potential energy function. This concept is often taught in the context of the Fundamental Theorem of Calculus, which states that the line integral of a gradient field is path-independent. In 1D, this is always true. A common misconception is that only linear or specific forces are conservative. This question tests the general definition and understanding of conservative forces. The distractors are meant to test if students mistakenly associate 'conservative' with a specific functional form.

📖 Explanation: The work done is the integral of the force. The average value of a function is defined as $\frac{1}{b-a} \int_a^b F(x) dx$. If the average force is 10 N, then $\frac{1}{5} \int_0^5 F(x) dx = 10$, so $\int_0^5 F(x) dx = 50$ J. Therefore, the work done must be 50 J. This is a direct Medium of the Mean Value Theorem for Integrals. The correct statement is that the work is exactly 50 J. This is a conceptual distinction between the average of a function and the average of its values at endpoints. Option A is the correct answer. The question is designed to test if students understand that the work is the integral, and the average force *defines* the value of the integral. Option B and C are distractors for those who think the average only applies to linear or constant functions.

📖 Explanation: This is a direct Medium of the work integral. $W = \int_0^2 2x dx = [x^2]_0^2 = 4$ J. It is a straightforward calculation. The distractors are common arithmetic errors. Option B is for those who integrate incorrectly (like forgetting the 2). Option C is for those who evaluate at x=1. Option D is for those who multiply incorrectly.

📖 Explanation: This is a critical Easy question. For a spring force $F=kx$, work $W = 0.5 k x^2$. If $x$ doubles, work quadruples. The student's reasoning is a classic mistake: confusing the linear relationship of force with the quadratic relationship of work. The correct answer is 'No, work is proportional to the square of extension'. Option C is a distractor that suggests a partially correct understanding (force times displacement) but misses the factor of 1/2 and the integration. Option D is another common misconception. The error lies in forgetting to integrate the force over the displacement, which is not a simple multiplication.

📖 Explanation: The correct integral for work is $\int_{x_0}^{x_1} F(x) \, dx$. So for $F(x)=3x$, it is $\int_0^2 3x \, dx$. Option B is the integral for a constant force of 3 N. Option C is the integral for a force $3x^2$. Option D incorrectly incorporates a factor of 2. This is a foundational recall question. The distractors target common mistakes in setting up the integral.

📖 Explanation: The graph is a line through the origin with slope 3, so the force is $F(x) = 3x$. The work is the area under the curve, which is a triangle with base 2 and height 6 (since $F(2)=6$). Area = $0.5 \times 2 \times 6 = 6$ J. Alternatively, $\int_0^2 3x dx = [1.5x^2]_0^2 = 6$. This tests the ability to interpret a graph as a function and calculate the integral geometrically. Option B is a distractor for those who take the average height (3) times the width (2). Option C is a distractor for those who integrate incorrectly or calculate the area of a rectangle.

📖 Explanation: $W = \int_0^{\pi/2} 2\sin x dx = [-2\cos x]_0^{\pi/2} = -2\cos(\pi/2) + 2\cos(0) = 0 + 2 = 2$ J. The distractors are common integration errors. Option B is for those who forget the factor of 2. Option C is for those who confuse the integral of sine with cosine. Option D is for those who incorrectly evaluate the limits.

📖 Explanation: The work is $W = \int x^n dx = \frac{x^{n+1}}{n+1}$ for $n \neq -1$. For $n = -1$, $W = \int x^{-1} dx = \ln x$. This question tests knowledge of special cases in integration. It is a multi-step reasoning problem: a student must know the general integration power rule, recognize the exception, and understand that this exception leads to a logarithmic function. Option B leads to a linear function, Option C to a quadratic, and Option D to a cubic. This question is designed to be a 5% Hard question.

📖 Explanation: The work is $W = \int_a^b [f(x) + c] dx = \int_a^b f(x) dx + c \int_a^b dx = \int_a^b f(x) dx + c(b-a)$. This combines the concept of linearity of integration with the definition of work. A common mistake is to treat the constant $c$ as a separate work term without multiplying by the displacement. Option B is a trap for forgetting the limits. Option D is a trap for an incorrect sign.

📖 Explanation: If $F(x)$ is scaled by 0.5, the new work is W&#039; = \int_1^4 0.5 F(x) dx = 0.5 \int_1^4 F(x) dx = 0.5 \times 30 = 15 J. This tests the linearity of the integral with respect to scalar multiplication. Option B is a common error for those who incorrectly multiply by 2. Option C is for those who think scaling the force doesn't change the work.

📖 Explanation: For an odd function like $F(x)=6x$, the integral from $-a$ to 0 is equal to the integral from 0 to $a$. $\int_{-2}^0 6x dx = [3x^2]_{-2}^0 = 0 - 12 = -12$ J. The work done is -12 J. This tests the symmetry properties of odd functions in integration. Option A is a trap for those who think the work is positive regardless of direction. Option C is for those who think the net displacement is zero. Option D is a distractor.

📖 Explanation: Work is the area under the force-displacement graph. For the first segment: $4 \times 2 = 8$ J. For the second segment: $6 \times 1 = 6$ J. Total work = $8 + 6 = 14$ J. This is a classic step function scenario. A common mistake is to take the average of the forces (5) and multiply by the total distance (3) to get 15 J, which is incorrect because the average force over the entire distance is not the same as the average of the two force values unless the intervals are equal. This question tests the understanding of integration as summation over piecewise functions.

📖 Explanation: If force is proportional to the square of displacement, $F = kx^2$. Work is $W = \int_0^L kx^2 dx = kL^3/3$. The student's result $kL^2$ suggests they incorrectly treated the force as constant at $kL$ or simply multiplied force times displacement without integration. This is a classic error of not integrating a variable force. Option B is for those who think the student is correct. Option C is for those who incorrectly integrate. Option D is a vague distractor.

📖 Explanation: The work is $\int_0^1 a e^{bx} dx = \frac{a}{b} [e^{bx}]_0^1 = \frac{a}{b} (e^b - 1)$. This tests the integration of exponential functions. Option B is a common error for forgetting the $1/b$ factor. Option C is a trap for evaluating the integral incorrectly. Option D is a trap for multiplying by $b$ instead of dividing.

📖 Explanation: The fundamental definition of work on a graph is the area under the force-displacement curve. This is a Easy question. The distractors test the common confusion between work, slope, and derivative. Option B is incorrect because slope would give the rate of change of force, which is related to spring constant. Option C is incorrect because the derivative would give the rate of change of work with respect to displacement, which is the force itself (by the Fundamental Theorem of Calculus). This question reinforces the basic graphical interpretation of the integral.

📖 Explanation: $W = \int_1^3 mx dx = m [x^2/2]_1^3 = m(9/2 - 1/2) = m(8/2) = 4m$. Since $W = 20$, $4m = 20$, so $m = 5$. This is a reverse Medium of the work integral. Students often solve for the variable incorrectly. Option B is the correct answer. Option A is a distractor for those who do the algebra wrong. Option C is a trap for those who think the work equals the force times displacement.

📖 Explanation: This question tests the fundamental reason for using calculus in work problems. Work is $F \times d$ only when the force is constant. When force varies with position, the total work is the sum of infinitesimal amounts of work $F(x) dx$, which is an integral. Option A is the precise reason. Option B is incorrect and suggests a misunderstanding of the concept. Option C is partially true but not the root cause; displacement is always a scalar quantity, and its linearity isn't the issue. Option D is a consequence of the definition, not the reason. This is a higher-order conceptual question that separates students who memorize formulas from those who understand the underlying physics.