Analyze Force Interactions Using Data - Grade 11 Physics

1. What vertical force does the hinge exert on the beam?
   Answer: $220 N$ upward, because torque balance about the hinge gives a cable tension of $280 N$, and vertical force equilibrium then requires the hinge to supply the remaining upward force. When a beam is in static equilibrium, two conditions must both be satisfied: the net torque about any point is zero, and the net force in every direction is zero. Questions like this one require you to apply both conditions in sequence — torque balance first, then force balance. Start by summing torques about the hinge (this eliminates the unknown hinge force from the equation). The beam weighs $200\,N$ acting at its center $(2.5\,m)$, and the person weighs $300\,N$ acting at $3.0\,m$. The cable pulls upward at the far end $(5.0\,m)$. Setting net torque to zero: $T(5.0) = 200(2.5) + 300(3.0)$ $T(5.0) = 500 + 900 = 1400$ $T = 280\,N$ Now apply vertical force equilibrium. The total downward force is $200 + 300 = 500\,N$. The cable provides $280\,N$ upward, so the hinge must supply: $F_{\text{hinge}} = 500 - 280 = 220\,N \text{ upward}$ That confirms C is correct. Choice A wrongly assumes the cable contributes nothing — it clearly does, since it's attached and pulling upward. Choice B makes a critical error: torque balance about the hinge gives the cable tension, not the hinge force (the hinge force produces no torque about that point, so it doesn't appear in that equation). Choice D invents a relationship between the person's weight and beam's weight that has no physical basis. The key strategy: in any static equilibrium problem with multiple unknowns, choose your torque pivot point to eliminate the force you don't want to find first, then use force balance to find what remains.
2. What is the cart's speed when the spring first reaches its uncompressed length?
   Answer: $2.37 m/s$, after subtracting the work done by friction from the spring energy. When a spring launches a cart on a surface with friction, you need the work-energy theorem: the net work done on the cart equals its change in kinetic energy. This means you must account for both the spring's stored energy and the energy lost to friction — not just one or the other. Start by finding the spring constant. Hooke's law gives $k = F/x = 12\text{ N} / 0.10\text{ m} = 120\text{ N/m}$. The elastic potential energy stored at compression $x = 0.15\text{ m}$ is $U_s = \frac{1}{2}kx^2 = \frac{1}{2}(120)(0.15)^2 = 1.35\text{ J}$. The work done against friction over that same 0.15 m distance is $W_f = (1.5)(0.15) = 0.225\text{ J}$. Applying the work-energy theorem: $\frac{1}{2}mv^2 = 1.35 - 0.225 = 1.125\text{ J}$, giving $v = \sqrt{2(1.125)/0.40} \approx 2.37\text{ m/s}$. That confirms A is correct. B ignores friction entirely, converting all 1.35 J into kinetic energy and getting $v \approx 2.60\text{ m/s}$ — a classic trap when students forget energy is always lost to friction before asking "where did it go?" C misuses the measured 12 N force as if it were constant throughout compression, computing work as $F \cdot d$ instead of using $\frac{1}{2}kx^2$. Since spring force varies linearly, you must use the energy formula. D confuses force with energy entirely — dividing force by mass gives acceleration, not speed. Your key takeaway: whenever friction acts during spring release, always subtract $W_f = fd$ from $U_s$ before solving for speed.
3. What is the magnitude of the contact force exerted by the $4.0 kg$ block on the $6.0 kg$ block?
   Answer: $18 N$, because the contact force must overcome friction on the $6.0 kg$ block and also provide the net force needed to accelerate it. When two objects move together as a system, you can analyze each one individually using Newton's second law — this is the key to finding contact forces. Start by finding the system's acceleration, then isolate the block of interest. First, find the acceleration of the entire system. The net force on both blocks combined is $30\,N - 4\,N - 6\,N = 20\,N$, and the total mass is $10\,kg$, giving $a = 2\,m/s^2$. Now isolate the $6.0\,kg$ block. The only horizontal forces acting on it are the contact force $F_c$ pushing it right and friction ($6\,N$) pushing it left. Applying Newton's second law: $F_c - 6\,N = (6.0\,kg)(2\,m/s^2) = 12\,N$ $F_c = 18\,N$ This confirms A is correct: the contact force must both overcome friction on the $6.0\,kg$ block and provide the net force needed to accelerate it. B is wrong because it ignores friction on the $6.0\,kg$ block entirely, only calculating $ma = 12\,N$. Friction is a real force that the contact force must also push against. C is wrong because $20\,N$ is the net force on the entire system — you cannot assign the full system's net force to a single contact point between the blocks. D is wrong because subtracting only the first block's friction ($30 - 4 = 26\,N$... not even $24\,N$) confuses the applied force with the contact force, ignoring that the first block also needs to accelerate. Study tip: When finding a contact force, always isolate one object and apply $F_{net} = ma$ to it alone — list every force on that object only, including friction and the unknown contact force.
4. What hanging mass most nearly produced the wave speed in the second trial?
   Answer: $0.46 kg$, obtained by relating the squared speed ratio to the tension ratio. Whenever you see a question linking wave speed to tension, remember the key relationship: wave speed is proportional to the square root of tension, meaning $v \propto \sqrt{T}$. This means if you want to find a tension ratio, you must work with the square of the speed ratio — not the speeds themselves. Here's how to solve it correctly (answer C). Start by writing the ratio of wave speeds: $\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}}$ Squaring both sides gives: $\frac{T_2}{T_1} = \left(\frac{v_2}{v_1}\right)^2 = \left(\frac{30}{40}\right)^2 = \frac{900}{1600} = 0.5625$ So $T_2 = 0.5625 \times 8.0\,\text{N} = 4.5\,\text{N}$. Since this tension equals the weight of the hanging mass, $m = \frac{T_2}{g} = \frac{4.5}{9.8} \approx 0.46\,\text{kg}$. Now for the distractors. Choice A applies the speed ratio twice (squaring the ratio, then applying it again to the mass), which is an extra, unjustified step. Choice B assumes tension is directly proportional to speed — ignoring the square root relationship entirely — giving an incorrect linear scaling. Choice D skips the ratio calculation altogether and uses the original 8.0 N tension to find mass, producing a result tied to the first trial, not the second. Your key takeaway: whenever $v \propto \sqrt{T}$, always square the speed ratio to get the tension ratio. Forgetting that squaring step is the most common trap on wave mechanics problems.
5. What is the approximate charge on each sphere?
   Answer: $1.42(10^{-6}) C$, using horizontal and vertical force equilibrium with Coulomb's law. When you see charged objects hanging from strings in equilibrium, your instinct should be to draw a free-body diagram and apply Newton's first law in both directions. Each sphere experiences three forces: gravity downward ($mg$), string tension along the string, and electrostatic repulsion horizontally. For equilibrium, dividing the horizontal equation by the vertical gives $\tan\theta = F_E / mg$, so $F_E = mg\tan(30°)$. Plugging in: $F_E = (0.020)(9.8)\tan(30°) = 0.196 \times 0.5774 \approx 0.1132 \text{ N}$. Then applying Coulomb's law: $F_E = kq^2/r^2$, so $q = \sqrt{F_E r^2 / k} = \sqrt{(0.1132)(0.40)^2 / 9.0\times10^9} \approx 1.42\times10^{-6} \text{ C}$. This confirms D is correct. Choice A makes the mistake of using only the horizontal component of weight ($mg\sin\theta$) instead of the proper ratio $mg\tan\theta$, which comes from combining both equilibrium equations correctly. This underestimates the electrostatic force. Choice B sets $F_E = mg$ (the full weight), ignoring that only the horizontal component of tension balances the electrostatic force — the full weight is balanced by the vertical tension component. Choice C uses $r$ instead of $r^2$ in Coulomb's law, a careless algebraic error that inflates the calculated charge significantly. Your key strategy: always resolve all forces into components before setting up equilibrium equations. The $\tan\theta$ relationship is your shortcut here — it appears whenever you divide horizontal by vertical equilibrium. Memorize it for pendulum and hanging-charge problems.
6. What is the crate's acceleration during the second trial?
   Answer: $2.0 m/s^2$, because the second trial has a net force of $40 N$. Whenever you see a question involving friction and Newton's Second Law, your first move should be to identify the net force — not just the applied force. Acceleration depends on the difference between applied force and friction, not on either value alone. Here's how to work through this problem. In the first trial, the crate moves at constant velocity, which means zero acceleration, which means zero net force. That tells you friction exactly equals the applied force: $f_k = 70 \text{ N}$. This is your key piece of information to carry into the second trial. In the second trial, the applied force is $110 \text{ N}$, but friction still opposes motion at $70 \text{ N}$. The net force is therefore $110 - 70 = 40 \text{ N}$. Applying Newton's Second Law: $a = \frac{F_{net}}{m} = \frac{40 \text{ N}}{20 \text{ kg}} = 2.0 \text{ m/s}^2$. That confirms C is correct. Choice A is wrong because kinetic friction is constant — it doesn't "adjust" to match whatever force is applied. That's a property of static friction up to its maximum value, not kinetic friction. Choice B misuses the first trial's applied force (70 N) as if it's relevant to the second scenario — it isn't, except to establish friction. Choice D ignores friction entirely and divides the full 110 N by the mass, which would only be valid if there were no friction at all. Study tip: When a problem says an object moves at constant velocity, treat that as a free gift — it tells you the net force is zero, letting you calculate friction directly. Always find friction first, then tackle the new scenario.
7. Approximately how far does the car travel while slowing from $10 m/s$ to rest in the later interval?
   Answer: $16.2 m$, using the braking force plus the later resistance. When a question gives you information from one scenario to apply to another, your job is to extract the right physics from the first scenario before solving the second. Here, the key is finding the braking force from the first interval, then combining it correctly with the new resistance. In the first interval, the car decelerates from $20\ m/s$ to $10\ m/s$ in $3.0\ s$, giving $a = \frac{10-20}{3.0} = -3.33\ m/s^2$. Newton's second law says the net force is $F_{net} = ma = 1200 \times (-3.33) = -4000\ N$. Since air and rolling resistance already contribute $500\ N$, the braking force alone is $4000 - 500 = 3500\ N$. Now in the later interval, the total resisting force is $3500 + 200 = 3700\ N$, giving $a = \frac{-3700}{1200} = -3.083\ m/s^2$. Using kinematics with $v_0 = 10\ m/s$ and $v = 0$: $d = \frac{v^2 - v_0^2}{2a} = \frac{0 - 100}{2(-3.083)} \approx 16.2\ m$. That's answer D. Choice A uses the full $4000\ N$ net force from the first interval rather than isolating the braking force, which incorrectly folds in the old resistance. Choice B ignores all resistance forces entirely, using only the $3500\ N$ braking force — this underestimates total braking and paradoxically gives a longer stopping distance. Choice C uses the correct forces but makes an algebra error by dropping the factor of 2 in the kinematic equation, doubling the actual distance. Your strategy: always isolate individual forces before transplanting them to a new scenario, and double-check your kinematic formula — the $2a$ in the denominator is an easy term to drop under pressure.
8. Which statement best describes the student's acceleration at that instant?
   Answer: The acceleration is $1.7 m/s^2$ upward because the normal force exceeds the weight. Whenever you see an elevator-and-scale problem, your go-to tool is Newton's Second Law applied to the vertical direction: $\sum F = ma$. The two vertical forces acting on the student are the normal force $N$ (upward, equal to the scale reading) and weight $W$ (downward). The difference between these forces — not either one alone — determines the acceleration. Start by finding the student's weight: $W = mg = 60 \times 9.8 = 588 \text{ N}$. The scale reads $690 \text{ N}$, which is greater than the weight, so the net force points upward. Applying Newton's Second Law: $N - W = ma$, which gives $690 - 588 = 60a$, so $a = \frac{102}{60} \approx 1.7 \text{ m/s}^2$ upward. This confirms B — the acceleration is $1.7 \text{ m/s}^2$ upward because the normal force exceeds the student's weight. A gets the magnitude right but reverses the direction. Since $N > W$, the net force is upward, not downward. Downward acceleration would require $N < W$. C confuses the scale reading being nonzero with having an acceleration equal to $g$. An acceleration of $9.8 \text{ m/s}^2$ would only apply if the student were in free fall — completely the opposite scenario. D divides the full scale force ($690 \div 60 = 11.5$) by mass, ignoring that weight must be subtracted first. You must find the net force before dividing by mass. Study tip: Always subtract weight from the scale reading to get net force — never divide the raw scale reading directly by mass.
9. Which statement correctly compares the average interaction forces during the collision?
   Answer: The bumper exerts $100 N$ leftward on the cart, and the cart exerts $100 N$ rightward on the bumper. When a question asks you to compare forces during a collision, your two guiding principles are impulse-momentum theorem and Newton's Third Law. Calculate the force on the cart first, then apply Newton's Third Law to get the force on the bumper instantly. Using the impulse-momentum theorem, the net force on the cart equals the change in momentum divided by time. Taking rightward as positive, the cart's initial velocity is $+3.0\ m/s$ and final velocity is $-1.0\ m/s$. So: $F = \frac{m\Delta v}{\Delta t} = \frac{(0.50\ kg)(-1.0 - 3.0)\ m/s}{0.020\ s} = \frac{-2.0\ kg\cdot m/s}{0.020\ s} = -100\ N$ The negative sign means the bumper pushes the cart leftward at $100\ N$. By Newton's Third Law, the cart simultaneously pushes the bumper rightward at exactly $100\ N$. This confirms B is correct. A is wrong because $50\ N$ comes from using only the rebound speed ($1.0\ m/s$) instead of the full velocity change ($4.0\ m/s$) — a very common arithmetic trap. C describes the right magnitude but flips the direction on the cart's force. A leftward rebound means the bumper pushed leftward, not rightward. D is doubly wrong: the magnitude is $100\ N$, not $200\ N$, and Newton's Third Law guarantees both forces are equal in magnitude — one object can never exert a "smaller" force in a Third Law pair. Study tip: Always compute $\Delta v = v_f - v_i$ with signs before plugging into impulse equations, and remember that Third Law pairs are always equal in magnitude, opposite in direction — no exceptions.
10. What is the skydiver's approximate speed at the end of the $0.50 s$ interval?
    Answer: $18.9 m/s$ downward, because the average net force produces an upward impulse. When a force changes over time, the impulse-momentum theorem is your go-to framework: $J = \Delta p = m\Delta v$. The impulse equals the net force multiplied by the time interval — but when force varies uniformly, you use the average net force. Here, the drag force increases linearly from $600\,N$ to $1200\,N$, so the average drag force is $\frac{600 + 1200}{2} = 900\,N$ upward. Weight is $735\,N$ downward. Taking downward as positive, the average net force is $735 - 900 = -165\,N$ (upward). The impulse over $0.50\,s$ is $J = (-165)(0.50) = -82.5\,N{\cdot}s$. Applying the impulse-momentum theorem: $\Delta v = \frac{J}{m} = \frac{-82.5}{75} \approx -1.1\,m/s$. Starting at $+20\,m/s$, the final speed is $20 - 1.1 \approx 18.9\,m/s$ downward — confirming B. A is wrong because the average drag ($900\,N$) exceeds weight ($735\,N$), so the net force is actually upward, not downward. Momentum decreases, not increases. C uses the final drag value ($1200\,N$) as if it acted the entire interval — this overstates the impulse. You must average a uniformly changing force, not use its endpoint. D ignores weight entirely when calculating the net impulse. Both forces act simultaneously; you must find the net force before multiplying by time. Study tip: On impulse problems, always check two things: (1) is the force changing — if so, average it; and (2) are multiple forces acting — if so, combine them into a net force before calculating impulse.

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