Statics and Dynamics Quiz: Work Energy Principle

What this quiz covers

This quiz focuses on Work Energy Principle, giving you a quick way to practice the rules, question types, and explanations that matter most for Statics and Dynamics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. $\tfrac{1}{2}(2m)v^2 = mgd$, because the work–energy principle applied to the system only tracks the kinetic energy of the particle that is acted upon by the net external force (particle B), while particle A's kinetic energy is accounted for separately through the cord tension.
2. $\tfrac{1}{2}(3m)v^2 = mgd - 2mgd$, because gravity acts on both particles: B gains kinetic energy from its descent while A loses potential energy as it moves horizontally, requiring both gravitational work terms to be included with opposite signs.
3. $\tfrac{1}{2}(3m)v^2 = mgd$, because the only external force doing work on the system is gravity acting on B (the table's normal force and the cord tension are internal or do no work), and both particles share the same speed due to the inextensible cord. (correct answer)
4. $\tfrac{1}{2}(3m)v^2 = mgd - T_{cord}\cdot d$, because the cord tension is an internal force that transmits energy between particles and must be subtracted from the gravitational work on B to correctly account for the energy transferred to particle A.

Question 2

1. $v \approx 6.81 \text{ m/s}$, from $\tfrac{1}{2}mv^2 = T(5) - mg\sin30°(5) - \mu_k mg(5)$, where the gravity component along the incline is handled correctly but the friction force is computed using the full weight rather than the normal force component on the incline.
2. $v \approx 8.42 \text{ m/s}$, from $\tfrac{1}{2}mv^2 = T(5) - mg\sin30°(5)$, accounting for tension and the gravity component along the incline but neglecting the friction force since the crate is moving uphill.
3. $v \approx 0.81 \text{ m/s}$, from $\tfrac{1}{2}mv^2 = T(5) - mg(5) - \mu_k mg\cos30°(5)$, where the full weight $mg$ (not just its incline component) is used to compute gravity's work and friction is computed using the correct normal force.
4. $v \approx 7.05 \text{ m/s}$, from $\tfrac{1}{2}mv^2 = T(5) - mg\sin30°(5) - \mu_k mg\cos30°(5)$, correctly accounting for tension work, the gravity component along the incline, and friction over 5 m. (correct answer)

Question 3

1. The student's error is using $v_1$ in the friction formula instead of $v_2$; the correct approach is to evaluate friction at the average speed $\bar{v} = (v_1+v_2)/2$ and iterate to convergence, which gives a higher $v_2$ than the student's result since friction was overestimated.
2. The student's error is treating the velocity-dependent friction force as constant at its initial value; since the friction force varies with speed, the work must be computed as $W_f = -\int_0^3 f\,dx$, which requires expressing $f$ as a function of position (via the equation of motion) rather than directly integrating over $x$, making the work–energy approach non-trivial for this force law. (correct answer)
3. The student's error is a sign convention mistake; the friction work should be positive ($+66 \text{ J}$) since friction acts opposite to displacement and the negative sign is already embedded in the force law, so the correct equation is $\tfrac{1}{2}(4)(8)^2 + 66 = \tfrac{1}{2}(4)v_2^2$, yielding a higher $v_2$.
4. The student's error is neglecting that the work–energy principle is invalid when non-conservative forces are velocity-dependent; only conservative forces can be handled by the work–energy principle, and velocity-dependent friction requires the impulse-momentum theorem instead.

Question 4

1. The cord does not break. The work–energy principle gives $v_{top}^2 = v_{bottom}^2 - 4gL = 36 - 4(9.81)(1.2) \approx -11.1 \text{ m}^2/\text{s}^2 < 0$, confirming the ball cannot reach the top. However, checking the cord tension at every intermediate angle requires a separate Newton's second law analysis along the radial direction, which is beyond what the work–energy principle alone provides.
2. The cord goes slack before the ball reaches the top. The work–energy principle gives $v_{top}^2 = v_{bottom}^2 - 4gL \approx -11.1 \text{ m}^2/\text{s}^2 < 0$, which is impossible, meaning the ball loses all kinetic energy at some angle before the top. At that point the speed is zero, the cord tension drops to zero and the cord goes slack — so the cord cannot break by snapping since the tension vanishes before it can reach $T_{max}$.
3. The cord does not break. The work–energy principle gives $v_{top}^2 = v_{bottom}^2 - 2gL = 36 - 2(9.81)(1.2) \approx 12.5 \text{ m}^2/\text{s}^2 > 0$, so the ball reaches the top. The tension at the top is $T_{top} = m(v_{top}^2/L) - mg \approx 0.28 \text{ N} < 30 \text{ N}$, so the cord does not break at the top; since the tension at the top is the minimum during circular motion, it does not break anywhere.
4. The cord does not break. The work–energy principle gives $v_{top}^2 = v_{bottom}^2 - 4gL \approx -11.1 \text{ m}^2/\text{s}^2 < 0$, confirming the ball cannot reach the top. The cord is most likely to break at the bottom, where tension is maximum. At the bottom: $T_{bottom} = m\left(\frac{v_{bottom}^2}{L} + g\right) = 0.5\left(\frac{36}{1.2} + 9.81\right) = 0.5(30 + 9.81) = 19.9 \text{ N} < 30 \text{ N}$, so the cord does not break at the bottom. Since this is the maximum tension during the motion, the cord does not break anywhere. (correct answer)

Question 5

1. $0 + mg(3) - \mu_k mg(d + 0.2) - \tfrac{1}{2}k(0.2)^2 = 0$, because all work terms — gravity, friction over the full contact distance including spring compression, and spring elastic potential — must sum to zero when the final kinetic energy is zero. (correct answer)
2. $0 + mg(3) - \mu_k mg \cdot d - \tfrac{1}{2}k(0.2)^2 = 0$, because friction acts only over the flat section of length $d$ and the spring does negative work equal to its stored elastic energy on the particle.
3. $0 + mg(3) - \mu_k mg(d + 0.2) = \tfrac{1}{2}k(0.2)^2$, because friction acts over distance $d + 0.2$ while the spring's elastic energy appears as positive kinetic energy gain rather than negative work on the particle.
4. $0 + mg(3) - \mu_k mg \cdot d = 0$, because the spring compression stores energy that exactly cancels the spring work term, leaving only gravity and friction in the work–energy balance at the stopping point.

Question 6

1. $v = \sqrt{\frac{2}{m}\left[\int_0^4(10+3x)\,dx + mg(0.5)\right]} \approx 8.6 \text{ m/s}$, obtained by integrating the variable force over 4 m and adding the gravitational work through the 0.5 m drop; the additional 2 m on the lower frictionless surface contributes no work since no force acts horizontally there. (correct answer)
2. $v = \sqrt{\frac{2}{m}\left[\int_0^6(10+3x)\,dx + mg(0.5)\right]} \approx 11.3 \text{ m/s}$, obtained by integrating the variable force over the full 6 m path (4 m upper + 2 m lower) and adding the gravitational work through the 0.5 m drop.
3. $v = \sqrt{\frac{2}{m}\int_0^4(10+3x)\,dx} \approx 8.0 \text{ m/s}$, obtained by integrating the variable force over 4 m only, since the gravitational work during the drop is a conservative internal effect that does not contribute to the work–energy calculation.
4. $v = \sqrt{\frac{2}{m}\left[F_{avg}(6) + mg(0.5)\right]} \approx 10.1 \text{ m/s}$, obtained by using the average force value $F_{avg} = 10 + 3(3) = 19 \text{ N}$ over the full 6 m path and adding gravitational work, treating the applied force as acting over the entire distance traveled.

Question 7

1. Student 2 is correct. Although drag does negative work per unit path length, it alters the trajectory so that the particle descends a greater vertical distance than $H$, increasing $W_{gravity}$ above $mgH$ and potentially yielding a higher landing speed than in vacuum.
2. Student 1 is correct for the given scenario, but Student 2's reasoning is partially valid: drag does alter the trajectory and increase path length, which means drag acts over a longer distance and removes even more energy than if the trajectory were unchanged, further reducing the landing speed below the vacuum value.
3. Student 1 is correct. By the work–energy principle, $T_{landing} = T_0 + W_{gravity} + W_{drag}$. Since gravity is conservative, $W_{gravity} = mgH$ regardless of the path taken — it depends only on the fixed elevation drop $H$. Since $W_{drag} < 0$ always (drag opposes velocity at every instant), the landing kinetic energy with drag is strictly less than without drag. (correct answer)
4. Neither student is definitively correct. The work–energy principle shows $W_{drag} < 0$ reduces kinetic energy, but the change in trajectory means the effective height drop may differ from $H$, making the comparison indeterminate without knowing the exact flight path.

Question 8

1. The equation is incorrect because gravity's work should appear on the left side as $-mg\Delta h$ added to $T_1$, not subtracted from $T_2$; moving it to the right side and treating it as a potential energy gain changes the form of the work–energy principle to an energy conservation form, which is only valid when $W_{nc} = 0$.
2. The equation is correct as written; it is a valid form of the work–energy principle where gravitational potential energy change $mg\Delta h$ appears on the right side to account for the energy stored against gravity, and $W_{nc}$ represents all non-conservative work inputs. (correct answer)
3. The equation is incorrect because the work–energy principle $T_1 + U_{1\to2} = T_2$ requires all work terms to appear on the left side; placing $mg\Delta h$ on the right side violates the standard statement of the theorem regardless of whether the result is algebraically equivalent.
4. The equation is incorrect because $W_{nc}$ should be replaced by the impulse of $\mathbf{F}_{nc}$; work and impulse are often confused in dynamics, and only the impulse-momentum theorem (not the work–energy principle) accommodates non-conservative forces correctly.

Question 9

1. Both the speed of the particle and the normal force exerted by the bowl on the particle at that point, because the work–energy principle together with geometry uniquely determines both the tangential velocity and the radial constraint force at any position on the bowl.
2. Only the speed of the particle, because the work–energy principle is a scalar equation relating kinetic energy changes to work done and yields speed directly; the normal force does no work and therefore does not appear in — and cannot be extracted from — the work–energy equation alone. (correct answer)
3. Only the normal force, because at any point on the bowl the geometry constrains the direction of the normal force, and the work–energy principle determines the energy state needed to compute it; speed requires Newton's second law in the tangential direction.
4. Neither the speed nor the normal force from work–energy alone, because the bowl is curved and the work–energy principle only applies to straight-line motion; curved-path problems require the tangential and normal equations of motion from Newton's second law.