2.5 Newton’s Second Law

1. AA cart rolling along a level floor gradually slows to a stop. Correct
   The cart's velocity is changing, since its speed decreases, and a system's velocity changes only if the net force on it is not zero. The backward horizontal forces on the cart from the floor and the air are not balanced by any forward force, so the forces are unbalanced.
2. BAn elevator carrying passengers rises at a steady speed.
   A student who thinks an object needs a net force in its direction of motion to keep moving picks this. The elevator's velocity is constant, so the net force on it is zero: the upward force exerted by the cable balances the downward gravitational force.
3. CA puck glides at constant velocity after a stick has hit it.
   A student who thinks the hit gives the puck a force that it carries along picks this. The stick exerts a force only while it touches the puck. Afterwards the puck's velocity is constant, so the forces on it are balanced.
4. DA heavy crate stays at rest while a student pushes on it gently.
   A student who thinks a push must exceed some threshold before it can move an object picks this, reasoning that the push acts but is 'not enough'. The crate's velocity stays zero, so the net force on it is zero: the floor exerts a horizontal force on the crate that balances the push.

CED 2.5.A.1 · Read this in Fix

1. AIts speed is increasing, as there is a net force exerted on it.
   A student who thinks a net force always speeds an object up picks this. A net force changes the velocity in the direction of the force: here the net force is to the right and the crate moves to the left, so it slows down.
2. BIts acceleration is to the left, the direction it is moving.
   A student who thinks acceleration always points in the direction of motion picks this. The acceleration has the direction of the net force, 10 N to the right, whichever way the crate is moving.
3. CIts speed is decreasing, as its acceleration is to the right. Correct
   The net force is 30 N − 20 N = 10 N to the right, so by Newton's second law the crate's acceleration is to the right. The crate is moving to the left, opposite to its acceleration, so its speed is decreasing.
4. DIts speed is constant, as the net force exerted on it is constant.
   A student who thinks a constant force produces a constant speed picks this. A constant net force produces a constant acceleration, so the velocity changes steadily; here the acceleration is opposite to the motion, so the speed decreases.

CED 2.5.A.2 · Read this in Fix

1. ANeither can move, since their center of mass must stay at rest.
   A student who thinks that what is true of the center of mass must be true of each part picks this. The center of mass stays at rest, but each student has an unbalanced force exerted on them by the other, so each one's velocity changes.
2. BThey move apart at equal speeds, and their center of mass stays at rest. Correct
   The pushes between A and B are internal forces of the two-student system, and the net external force is zero, so the velocity of the center of mass stays zero. The pushes are equal in magnitude and last equally long, so, with equal total masses, they move apart at equal speeds.
3. CB moves off faster than A, since A pushes on B harder than B pushes on A.
   A student who thinks the object doing the pushing exerts the larger force picks this. B pushes back on A with a force of equal magnitude for the same time, so with equal masses the two speeds are equal.
4. DThey move apart, and their center of mass moves the way A pushes.
   A student who thinks a push between parts of a system can move the whole system picks this. A's push on B and B's push on A are internal to the system and add to zero, so the center of mass stays at rest.

CED 2.5.A.3 · Read this in Fix

1. A32 N
   A student who thinks the net force is set by the velocity multiplies the mass by the final velocity: (8.0 kg)(4.0 m/s) = 32 N. The net force determines how fast the velocity changes, which is the slope of the graph, not its height.
2. B16 N
   A student who divides the final velocity by the time gets a = (4.0 m/s)/(2.0 s) = 2.0 m/s² and F = 16 N. The cart did not start from rest, so the change in velocity, 3.0 m/s, must be used: a = 1.5 m/s².
3. C40 N
   A student who takes the area under the graph as the acceleration uses the area, 5.0 m, and multiplies by the mass to get 40 N. The area is the cart's displacement; the acceleration is the slope, 1.5 m/s².
4. D12 N Correct
   The acceleration is the slope of the graph, (4.0 m/s − 1.0 m/s)/(2.0 s) = 1.5 m/s². By Newton's second law, F_net = ma = (8.0 kg)(1.5 m/s²) = 12 N.

Working The acceleration is the slope of the velocity–time graph: a = Δv/Δt = (4.0 m/s − 1.0 m/s)/(2.0 s) = 1.5 m/s². Newton's second law: F_net = m a = (8.0 kg)(1.5 m/s²) = 12 N.

CED 2.5.A.2 · Read this in Fix

1. AT = F
   A student who thinks the string passes on the whole of the student's pull picks this. Part of F is needed to accelerate block P, so the string pulls Q with less than F: T = 2F/3.
2. BT = 2F/3 Correct
   Treat both blocks as one system of mass 3m: the only horizontal external force is F, so a = F/(3m). The string is the only horizontal force on Q, so T = (2m)(F/3m) = 2F/3.
3. CT = 2F
   A student who divides F by the mass of block P alone gets a = F/m and then T = (2m)(F/m) = 2F. The pull must accelerate both blocks, so a = F/(3m), and T = 2F/3, which is less than F.
4. DT = F/3
   A student who multiplies the correct acceleration, F/(3m), by the mass of block P gets F/3. The string pulls block Q, so the tension equals Q's mass times the acceleration: (2m)(F/3m) = 2F/3.

Working System P + Q (mass 3m): the only horizontal external force is F, so a = F/(3m). Object Q: the only horizontal force exerted on Q is the tension, so T = (2m)a = (2m)(F/3m) = 2F/3. Check on P: F − T = m a gives F − 2F/3 = F/3 = m(F/3m).

CED 2.5.A.2 · Read this in Fix

1. AA straight line through the origin with half the slope Correct
   Newton's second law gives a = (1/m)F_net: a straight line through the origin with slope 1/m. Doubling the mass halves the slope, so every net force now produces half the acceleration it did before.
2. BA straight line through the origin with twice the slope
   A student who reads the slope of the graph as the mass picks this. Rearranged for these axes, the law is a = (1/m)F_net, so the slope is 1/m; doubling the mass halves the slope instead of doubling it.
3. CThe same line, since the net forces used are not changed
   A student who thinks acceleration depends only on the net force picks this. For the same net force, a system of twice the mass has half the acceleration, so every point moves down to half its height.
4. DA line of the same slope that crosses the F axis at a positive value
   A student who thinks a force must exceed a threshold before a more massive object will accelerate picks this. Any nonzero net force gives a nonzero acceleration, so the line still starts at the origin; only its slope changes.

Working a = F_net/m, so a graph of a against F_net is a straight line through the origin with slope 1/m. Here the slope is (2.0 m/s²)/(4.0 N) = 0.50 kg⁻¹, so m = 2.0 kg. Doubling the mass to 4.0 kg halves the slope to 0.25 kg⁻¹: each net force now gives half the acceleration, and the line still passes through the origin.

CED 2.5.A.2 · Read this in Fix

1. AThe net force exerted on each cart was negligible during the collision.
   A student who assumes each part of a system behaves like its center of mass picks this. Between 0.10 s and 0.20 s both carts' velocities change quickly, so a large net force was exerted on each cart; only the net external force on the whole system was negligible.
2. BCart 1 exerted a larger force on cart 2 than cart 2 exerted on cart 1.
   A student who thinks the moving object exerts the larger force picks this. Over the same 0.10 s, cart 1's velocity changes by 0.80 m/s (a = 8.0 m/s², force 8.0 N) and cart 2's by 0.40 m/s (a = 4.0 m/s², force 8.0 N): the forces are equal in magnitude.
3. CA forward net external force was needed to keep the system moving.
   A student who thinks a moving system needs a net force to keep moving picks this. The center of mass moves at a constant 0.20 m/s, which requires zero net external force, not a forward one.
4. DThe net external force on the two-cart system was negligible. Correct
   The velocity of the center of mass stays at 0.20 m/s throughout, including during the collision. The velocity of a system's center of mass changes only if a nonzero net external force is exerted on the system, so the net external force was negligible.

CED 2.5.A.3 · Read this in Fix

1. AIt is greater in case 2, since the hand then pushes the lighter block.
   A student who divides F by the mass of the block the hand touches gets F/(2m) in case 1 and F/m in case 2. Both blocks must be accelerated together, so the mass to use is the total, 3m, in both cases.
2. BIt is greater in case 1, since the blocks push on each other less then.
   A student who thinks the push between the blocks can slow the pair picks this. The contact force is indeed smaller in case 1 (F/3, against 2F/3 in case 2), but it is internal to the system and does not affect the system's acceleration.
3. CIt is the same in both cases, since the net external force is F in each. Correct
   In both cases the system is both blocks, of total mass 3m, and the only horizontal external force is the hand's push F. So a = F/(3m) in each case. Which block the hand touches changes only the internal contact force between the blocks.
4. DIt is zero in both cases, as the blocks' pushes on each other cancel.
   A student who thinks third-law pairs cancel picks this. A's push on B and B's push on A act on different blocks, and they are internal to the system in any case; the external push F gives the system an acceleration F/(3m).

CED 2.5.A.2 · Read this in Fix

1. A√(2Fd)
   A student who thinks the acceleration depends only on the net force, and not on the mass, uses a = F and picks this. The expression does not even have units of speed; the same net force gives a more massive cart a smaller acceleration, a = F/m.
2. B√(2Fd/m) Correct
   The acceleration is a = F/m, constant. From rest, v² = 2ad = 2Fd/m, so v = √(2Fd/m). A more massive cart has a smaller acceleration and so reaches a smaller speed over the same distance.
3. C√(Fd/m)
   A student who finds a = F/m correctly but uses v² = a d, without the factor 2, picks this. From rest, v² = 2ad = 2Fd/m, so the speed is √2 times larger.
4. D2Fd/m
   A student who reaches v² = 2Fd/m and stops there, without taking the square root, picks this. That expression is v², with units m²/s²; the speed is its square root.

Working Newton's second law: the cart's acceleration is a = F/m, constant and along the track. From rest, v_x² = v_x0² + 2a_x(x − x₀) = 2(F/m)d, so v = √(2Fd/m). Units: N·m/kg = m²/s², whose square root is m/s. (a taken as F, independent of mass: √(2Fd), not a speed. Factor 2 dropped, v² = a d: √(Fd/m). Square root omitted: 2Fd/m, which is v², in m²/s².)

CED 2.5.A.2 · Read this in Fix