📝 Work energy theorem calculus (37 MCQs)

📖 Explanation: This question requires applying the work-energy theorem where work done by a variable force is $W = \int F(x) dx$, and this work equals the change in kinetic energy. Computing $\int_0^3 4x dx = [2x^2]_0^3 = 18$ J. Since initial kinetic energy is zero, final kinetic energy is 18 J. Distractor A is half the correct value; C uses the final position as the coefficient; D uses the average force incorrectly.

📖 Explanation: The work-energy theorem states $W = \Delta K = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2 = 0 - \frac{1}{2}(5)(4)^2 = -40$ J. Work is also $W = Fd\cos\theta$, and since the force opposes motion, $\theta = 180^\circ$, so $W = -Fd$. Thus $-F(8) = -40$, giving $F = 5$ N. Distractor A results from dividing initial kinetic energy by distance; B uses mass incorrectly; C comes from incorrect sign handling.

📖 Explanation: Work is $W = \int_0^4 (6 - 2x)dx = [6x - x^2]_0^4 = 24 - 16 = 8$ J. From work-energy theorem, $W = \Delta K = \frac{1}{2}mv^2 - 0$, so $8 = \frac{1}{2}(10)v^2$, giving $v = \sqrt{1.6} \approx 1.79$ m/s. Distractor C uses the average force incorrectly; D uses the final position; A would only occur if net work were zero.

📖 Explanation: The student's claim is incorrect because the work-energy theorem applies to the net work done by all forces. If a force does positive work but other forces do larger negative work, the net work could be negative, causing kinetic energy to decrease. Distractor A fails to consider net work; C incorrectly defines work; D misunderstands the directional nature of work while correctly noting that work is path-dependent but this isn't the primary reason the statement is false.

📖 Explanation: Work done by a variable force is $W = \int F(x) dx = \int_R^{2R} \frac{k}{x^2} dx = [-\frac{k}{x}]_R^{2R} = -\frac{k}{2R} + \frac{k}{R} = \frac{k}{2R}$. The negative sign in the antiderivative is important. Distractor A uses the value at the upper limit only; C uses the value at the lower limit incorrectly; D is double the correct value from a sign error in evaluating the definite integral.

📖 Explanation: Kinetic energy is $K = \frac{1}{2}mv^2$. If $K_1 = K_2$, then $\frac{1}{2}mv_1^2 = \frac{1}{2}(2m)v_2^2$, so $v_1^2 = 2v_2^2$, giving $v_1 = \sqrt{2}v_2$. Thus the lighter object (mass m) has $\sqrt{2}$ times the speed of the heavier object. Distractor A confuses mass and speed relationships; B inverts the ratio; D only occurs when masses are equal.

📖 Explanation: By definition, the work done by a variable force is the integral of force with respect to displacement, which is graphically represented as the area under the force-displacement curve. Distractor A confuses work with impulse, which is force integrated over time; C confuses work with power (work/time); D confuses work with impulse (force × time).

📖 Explanation: From the work-energy theorem in differential form, $dK = F(x)dx$, so $F(x) = dK/dx$. Given $K(x) = 6x^2$, $dK/dx = 12x$. Thus $F(x) = 12x$ N. This is a higher-order thinking question because it requires differentiating the kinetic energy function with respect to position. Distractor B is the original force that might have been integrated; C misapplies the derivative; D incorrectly multiplies the original expression.

📖 Explanation: The work-energy theorem states that the net work done by all forces equals the change in kinetic energy. If the net work is zero, the kinetic energy cannot change. However, individual forces could do work while others do opposite work, resulting in zero net work. The student's statement is correct in the sense that if a single force does no work, it doesn't directly change kinetic energy, but other forces could. Distractor B misrepresents internal forces; D contradicts Newton's laws; C is correct but needs the 'net' qualifier.

📖 Explanation: Work is $W = \int_0^5 (2x+1) dx = [x^2 + x]_0^5 = 25 + 5 = 30$ J. Initial kinetic energy $K_i = \frac{1}{2}(4)(3)^2 = 18$ J. From work-energy theorem, $K_f = K_i + W = 18 + 30 = 48$ J. Distractor A is the initial kinetic energy only; B is half the final energy due to a division error; D is the work done incorrectly computed as $\frac{1}{2}(5)(2x+1)$.

📖 Explanation: Work done is $W = \int_0^A -kx dx = -k[\frac{x^2}{2}]_0^A = -\frac{1}{2}kA^2$. The negative sign indicates that the force opposes the displacement (it's a restoring force like a spring). Distractor B misses the negative sign; C is the integral without the factor of 1/2; D has both sign and factor errors. This is a conceptual question testing the understanding that work can be negative and the integral of a linear force gives a quadratic dependence.

📖 Explanation: Kinetic energy $K = \frac{1}{2}mv^2$. If $K$ increases by a factor of 4, then 4K = \frac{1}{2}m(v')^2. Since $K = \frac{1}{2}mv^2$, we have 4(\frac{1}{2}mv^2) = \frac{1}{2}m(v')^2, so 4v^2 = (v')^2, giving v' = 2v. Distractor B confuses energy factor with speed factor; C squares the speed factor incorrectly; D incorrectly takes the square root of 4 as $\sqrt{2}$.

📖 Explanation: The kinetic energy of the particle is equal to the total work done by the force, which is the area under the force-displacement curve from 0 to the position. The kinetic energy will be maximum when the cumulative area under the curve is maximum, which occurs at position C where the force has just become zero after being positive for the entire interval. At D, the force is negative, so it does negative work, decreasing kinetic energy. Distractor A is where the force is maximum but not where cumulative area is maximum; B is before the cumulative area is maximum; D is where negative work has been done.

📖 Explanation: Work done is $W = \int_0^{\pi/2} 4\sin x dx = [-4\cos x]_0^{\pi/2} = -4(0) - (-4(1)) = 4$ J. From work-energy theorem, $K_f = K_i + W = 2 + 4 = 6$ J. Distractor A is the initial kinetic energy only; B is the work done; D is twice the work done. This question tests integration of trigonometric functions and Medium of the work-energy theorem.

📖 Explanation: From the work-energy theorem, $F(x) = dK/dx$. Differentiating $K(x) = 3x^3 - 2x^2 + 1$ gives $F(x) = 9x^2 - 4x$. At $x=2$, $F(2) = 9(4) - 4(2) = 36 - 8 = 28$ N. This is a Hard question combining calculus (differentiation) with physics (work-energy relationship). Distractor A is the value at $x=1$; C is the value at $x=0$; D uses $K(2)$ directly without differentiating.

📖 Explanation: The statement is false. An object can have zero kinetic energy (i.e., be at rest) while forces are acting on it, as long as the forces are balanced (net force is zero) or the object is at a turning point where forces do no work. More fundamentally, kinetic energy is a function of speed, not force. A net force can be zero while kinetic energy is zero (object at rest), but zero kinetic energy doesn't require zero net force - consider an object thrown upward at the highest point: kinetic energy is zero but net force (gravity) is not zero. Distractor A incorrectly links energy to motion; C misapplies conservation laws; D has the right answer but for the wrong reason.

📖 Explanation: Work done = change in kinetic energy = $\frac{1}{2}m(v_f^2 - v_i^2) = \frac{1}{2}(3)(2^2 - 8^2) = \frac{3}{2}(4 - 64) = \frac{3}{2}(-60) = -90$ J. The negative work means the net force opposed the motion. Distractor B has the wrong sign; C uses incorrect mass or velocity values; D has both magnitude and sign errors. This is a straightforward Medium of the work-energy theorem.

📖 Explanation: Since $F(x) = dK/dx$, the force is positive when the derivative of K is positive (kinetic energy increasing) and negative when the derivative is negative (kinetic energy decreasing). At the maximum of K, the derivative is zero, so the force changes sign from positive to negative. Distractor A is wrong because the force becomes negative when K decreases; C is wrong because the force is positive when K increases; D would result in K changing linearly, not increasing then decreasing.

📖 Explanation: Work done is $W = \int_0^2 (2x+3) dx = [x^2 + 3x]_0^2 = 4 + 6 = 10$ J. From work-energy theorem, $K_f = K_i + W = 5 + 10 = 15$ J. Distractor A comes from evaluating the integral only at the upper limit; B uses the integral without the lower limit; D is the result of adding the work to twice the initial energy.

📖 Explanation: The work done is $W = \Delta K = 60 - 20 = 40$ J. If the same force is applied over twice the distance, the work done is doubled (assuming constant force), so W' = 80 J. The new final kinetic energy is K_f' = K_i + W' = 20 + 80 = 100 J. Distractor B adds the original $\Delta K$ incorrectly; C adds 3 times the original $\Delta K$; D adds 4 times the original $\Delta K$.

📖 Explanation: Work done is $W = \int_0^5 3x^2 dx = [x^3]_0^5 = 125$ J. From work-energy theorem, $K_f = K_i + W = 10 + 125 = 135$ J. Distractor A is the work done if the force was $F(x) = 3x$ (incorrect integration); B is the work done if integrated as $\frac{3}{2}x^2$; C is the work done at $x=5$ but without adding the initial energy.

📖 Explanation: Kinetic energy $K = \frac{1}{2}mv^2$ and momentum $p = mv$. If kinetic energy doubles, 2K = \frac{1}{2}m(v')^2, so (v')^2 = 2v^2, giving v' = \sqrt{2}v. Therefore momentum p' = mv' = m(\sqrt{2}v) = \sqrt{2}p. Distractor A would require the speed to double; C would require the energy to quadruple; D would require the speed to remain constant.

📖 Explanation: Kinetic energy is the cumulative area under the force-displacement curve. From 0 to 4 m, the force is positive, so kinetic energy increases. From 4 to 6 m, the force is still positive but decreasing, so kinetic energy continues to increase but at a decreasing rate. From 6 to 8 m, the force is negative, so the force does negative work and kinetic energy decreases. Therefore, the maximum kinetic energy occurs at $x=6$ m, where the force changes sign from positive to negative. Distractor A is where the force is maximum but not where cumulative area is maximum; C is where net work might be zero if the areas under positive and negative parts are equal; D is incorrect because the force is not zero everywhere.

📖 Explanation: The force is $F(x) = dK/dx = 8x + 2$. At $x=3$, $F(3) = 8(3) + 2 = 26$ N. This is a Hard question requiring differentiation of the kinetic energy function. Distractor A is the derivative evaluated at $x=1$; B is the derivative at $x=2.5$; C is the derivative at $x=2.75$.

📖 Explanation: Initial kinetic energy $K_i = \frac{1}{2}(4)(5)^2 = 50$ J. Work done = 30 J, so $K_f = 50 + 30 = 80$ J. Then $80 = \frac{1}{2}(4)v_f^2 = 2v_f^2$, so $v_f = \sqrt{40}$ m/s. Distractor A uses $K_i = 20$ J (incorrect mass factor); B uses $K_i = 40$ J; C uses $K_f = 60$ J from adding work to 30 J instead of 50 J.

📖 Explanation: Net work is the sum of work done by each force: $W_1 = \int_0^3 4x dx = [2x^2]_0^3 = 18$ J. $W_2 = \int_0^3 -6 dx = [-6x]_0^3 = -18$ J. Net work $W_{net} = 18 - 18 = 0$ J. Distractor A would be correct if the forces cancel completely; B incorrectly sums the work values; C uses the force values at $x=3$ only; D is the work done by $F_1$ alone.

📖 Explanation: $F(x) = dK/dx = 4x + 3$. At $x=0$, $F(0) = 3$ N. Distractor A is the coefficient of $x^2$; C is the constant term in K(x); D is the sum of the coefficients. This tests understanding that force is the derivative of kinetic energy with respect to position.

📖 Explanation: Since $F(x) = dK/dx$, the force is zero when the derivative of K with respect to x is zero, which corresponds to the slope of the K versus x graph being zero. This occurs at maxima, minima, or inflection points where the tangent is horizontal. Distractor A describes where the magnitude of the force is maximum; C describes the minimum of K, which could have zero slope but not always; D describes a region where the force is zero but not a specific point.