Unit 2 Practice Set: Force and Translational Dynamics

(C). x_cm = (m₁x₁ + m₂x₂)/(m₁ + m₂) = (2.0·0 + 6.0·4.0)/(8.0) = 24/8 = 3.0 m.

(B) is the plain middle point. That would be right only if the masses were equal. The center of mass sits closer to the heavier ball. (D) puts it on the heavy ball itself, which ignores the light ball.

(B). The pushes are internal forces: each skater pushes the other. Internal forces cancel in pairs (third law). Gravity and the normal force from the ice also cancel. So the net external force is zero and the center of mass keeps its velocity, which was zero.

(A), (C) and (D) all need a net external force on the system. There is none.

(C). Up positive. The book is at rest, so net force = 0: N − 10 N − 5 N = 0, so N = 15 N.

(B) is the mistake "normal force always equals weight." The normal force is whatever the surface must push to stop the book from sinking into it. (A) uses only your push. (D) double counts.

(B). Vertical forces (up positive): N + 20 sin 30° − mg = 0. Weight mg = 5.0 · 9.8 = 49 N. The rope's up part is 20 · 0.50 = 10 N. So N = 49 − 10 = 39 N.

(C) ignores that the rope lifts a little. (D) adds the rope's pull instead of subtracting. (A) uses 20 cos 30° = 17.3 N, the horizontal part, which does not act up or down.

(C). Two forces act: tension up, weight down. At rest, T = mg = 3.0 · 9.8 = 29.4 N ≈ 29 N.

(D) confuses "no net force" with "no forces." (A) gives the mass, not a force. (B) is g, not a force.

(C). These two forces are a Newton's third law pair. A pair is always equal in size and opposite in direction, no matter the masses or speeds.

The car is damaged more and speeds up more because the same force acts on a smaller mass (a = F/m), not because the force is bigger. That is why (A) and (B) are wrong. (D) is wrong because forces always come in pairs.

(C). A third law pair swaps the two objects: "Earth pulls book" pairs with "book pulls Earth." Both are gravity.

(A) is equal and opposite here, but both forces act on the same object (the book) and are different types (normal and gravity). That is the first law balance, not a third law pair. (B) is the partner of the table's normal force. (D) is a real force, but it acts between the table and Earth, not between the book and Earth, so it cannot be the partner of W.

(B). Constant velocity means zero acceleration, so net force is zero (first law). Moving does not need a force; changing velocity does.

(A) and (C) are the old "motion needs a force" idea. (D) is wrong because gravity and the normal force still act. They just cancel.

(B). Constant velocity, so a = 0 and net force is 0. Only tension (up) and weight (down) act, so they are equal.

Tension is greater than weight only while the elevator speeds up going up (or slows down going down). The speed itself does not matter, so (D) is wrong.

(B). a = F_net/m = 12 N / 4.0 kg = 3.0 m/s².

(A) divides the wrong way. (C) subtracts. (D) multiplies.

(C). The scale reads the normal force N. Up positive: N − mg = ma, so N = m(g + a) = 60(9.8 + 2.0) = 60 · 11.8 = 708 N.

(B) is the reading at rest or at constant speed. (A) is the reading if the acceleration pointed down. (D) uses g = 10 m/s², which is close but not what we asked for here.

(B). Treat both blocks and the string as one system. The net force along the string is the difference in weights: (5.0 − 3.0)(9.8) = 19.6 N. The total mass moved is 8.0 kg. a = 19.6 / 8.0 = 2.45 m/s².

(C) divides 19.6 N by only 6 kg (wrong mass). (A) divides by 10 kg. (D) is free fall; the string slows both blocks.

(C). Both blocks move together: a = 18 N / 6.0 kg = 3.0 m/s². The only horizontal force on B is the push from A: F = m_Ba = 4.0 · 3.0 = 12 N.

(D) assumes your push passes through unchanged; part of it is used to speed up A. (A) is m_Aa, the net force on A. (B) just halves 18 N.

(C). A straight v-t line has a constant slope, so the acceleration is constant. The slope is negative, so the acceleration (and so the net force, F = ma) points in the negative direction, opposite to the motion.

(D) mixes up velocity and force: the velocity reaches zero, but the slope (and the force) stays the same. (A) would mean a flat line.

(B). Mass is the amount of matter; it does not change with place. Weight is the gravitational force: W = mg = 80 · 1.6 = 128 N.

(A) changes the mass. (C) uses Earth's g. (D) is the myth that there is no gravity on the Moon.

(C). F = Gm₁m₂/r². Double r and r² becomes 4 times bigger, so F becomes 1/4 as big.

(B) forgets the square. (A) has the wrong direction of change: farther means weaker.

(B). g = GM/R². Mass × 2 makes g twice as big. Radius × 2 makes R² four times as big, so g is 1/4. Together: 2/4 = 1/2. g = 9.8/2 = 4.9 m/s².

(C) thinks the two doublings cancel; they do not, because R is squared. (D) only uses the mass. (A) uses 1/4 only.

(C). The heavier ball does feel 5 times the force, but it also has 5 times the mass. a = F/m = (mg)/m = g for both, so they fall together.

(D) gets the result right but the reason wrong: the forces are not the same (9.8 N and 49 N). (A) and (B) each look at only half of F = ma.

(A). First check if it slides. N = mg = 98 N. The most static friction can give is μ_sN = 0.50 · 98 = 49 N. Your 30 N push is less than 49 N, so the box stays. Static friction matches your push: 30 N.

(C) treats μ_sN as the friction always; it is only the limit. (B) uses kinetic friction, but the box is not sliding. (D) is the normal force.

(B). 60 N is more than the 49 N static limit, so the box slides. Sliding means kinetic friction: f_k = 0.30 · 98 = 29.4 N. a = (60 − 29.4)/10 = 3.06 ≈ 3.1 m/s².

(A) uses static friction (49 N) for a sliding box. (D) ignores friction.

(C). Constant speed: net force = 0. Along the ramp: mg sin θ = μ_kN. Across the ramp: N = mg cos θ. So μ_k = sin θ / cos θ = tan 20° = 0.36.

(B) is sin 20°. (D) is cos 20°. (A) is the angle divided by 100, which has no meaning.

(A) and (B). Tilt the axes along the ramp. Across the ramp nothing moves: N = mg cos θ. Along the ramp friction holds the block: f_s = mg sin θ.

(C) is true only on level ground. (D) is the largest friction the ramp could give; it equals the real friction only when the block is just about to slip.

(C). Hooke's law: |F| = kΔx = 200 · 0.15 = 30 N, pulling back toward the natural length.

(B) and (D) divide instead of multiply. (A) divides the other way.

(C). At rest, spring force = weight: kΔx = mg. k = mg/Δx = (0.50 · 9.8)/0.049 = 4.9/0.049 = 100 N/m.

(A) uses the mass 0.50 kg instead of the weight 4.9 N. (B) divides 4.9 N by 0.10 m, twice the real stretch. (D) is off by a factor of 10.

(C). Parallel: each spring holds half the weight, so each stretches W/(2k). Total stretch W/(2k). Series: each spring holds the whole weight and stretches W/k, and there are two, so the total is 2W/k. Ratio (2W/k)/(W/2k) = 4.

(B) catches only one of the two effects. (D) flips the ratio.

(B). Static friction is the only force toward the center. At the limit: μ_smg = mv²/r, so v = √(μ_sgr) = √(0.80 · 9.8 · 50) = √392 = 19.8 m/s.

(D) forgets the square root. (A) leaves out g. (C) has no square root and a stray factor of 10.

(A). Once the string breaks, no horizontal force acts. By the first law the ball keeps its velocity at that instant, which points along the tangent: north.

(B) is the "centrifugal force" myth: nothing pushes the ball outward. (C) needs a force to keep bending the path. (D) is the direction of the old tension, not of the velocity.

(C). At the top, gravity and the normal force both point down, toward the center: N + mg = mv²/r. The smallest speed is when the track just barely touches, N = 0: v = √(gr) = √(9.8 · 10) = √98 = 9.9 m/s.

(A) forgets that the car must keep turning. (B) is √9.8: it leaves out the radius. (D) forgets the square root.

(D). At the bottom the center is straight up. Up positive: T − mg = mv²/r. mv²/r = 0.50 · 16 / 0.80 = 10 N. mg = 4.9 N. T = 10 + 4.9 = 14.9 N.

(C) is only the net force; it forgets that tension must also hold up the weight. (B) subtracts the weight (that would be right at the top). (A) is just the weight.

(C). Gravity provides the centripetal force: GMm/r² = mv²/r, so v = √(GM/r). With r × 4, v changes by 1/√4 = 1/2.

(D) uses 1/r² for the speed; that is how the force changes, not the speed. Farther orbits are slower, so (A) and (B) are wrong.

(a) Block A: weight m_Ag down (from Earth); normal force N up (from the table), same length as the weight; tension T to the right toward the pulley (from the string); kinetic friction f_k to the left (from the table), shorter than T. Block B: weight m_Bg down (from Earth); tension T up (from the string), shorter than the weight because B speeds up downward.

(b) Take the direction of motion as positive for each block (A to the right, B down). Block A: T − μ_km_Ag = m_Aa. Block B: m_Bg − T = m_Ba. Add them to cancel T: m_Bg − μ_km_Ag = (m_A + m_B)a, so

a = (m_B − μ_km_A) g / (m_A + m_B)

(c) a = (3.0 − 0.20 · 2.0)(9.8)/(5.0) = (2.6)(9.8)/5.0 = 25.48/5.0 = 5.1 m/s². T = m_B(g − a) = 3.0(9.8 − 5.10) = 14.1 N. Check with block A: T = μ_km_Ag + m_Aa = 3.92 + 10.19 = 14.1 N. Same.

(d) Constant acceleration from rest: v² = v₀² + 2aΔy = 0 + 2(5.10)(0.50) = 5.10 m²/s², so v = 2.3 m/s.

(e) Increases. From (b), a larger μ_k makes the top of the fraction smaller, so a is smaller. From block B's equation, T = m_B(g − a). A smaller a means a larger T. In words: B speeds up less, so the string must hold B back more.

Scoring (10 points)

• (a) 2 points: 1 point for block A with exactly four correct forces and directions (friction opposite to motion); 1 point for block B with T shorter than the weight. Minus 1 for any extra force (such as a "force of motion").
• (b) 3 points: 1 point for a correct second-law equation for A; 1 point for a correct one for B (consistent signs); 1 point for combining them to the final expression.
• (c) 2 points: 1 point for a = 5.1 m/s²; 1 point for T = 14 N (accept 14.1 N). Answers consistent with an error carried from (b) can earn these points.
• (d) 1 point for v = 2.3 m/s using a kinematics equation with the acceleration from (c).
• (e) 2 points: 1 point for "increases"; 1 point for linking smaller a to larger T through T = m_B(g − a). No credit for the claim alone.

(a) Two forces: the normal force N from the scale, up; the weight mg from Earth, down. N is longer than mg, because the elevator speeds up while moving up, so the net force points up.

(b) The accelerations are the slopes of v-t: 0 to 2 s: a = (4.0 − 0)/2 = +2.0 m/s². 2 to 8 s: a = 0. 8 to 10 s: a = (0 − 4.0)/2 = −2.0 m/s². From N − mg = ma:

• 0 to 2 s: N = 50(9.8 + 2.0) = 590 N
• 2 to 8 s: N = 50(9.8) = 490 N
• 8 to 10 s: N = 50(9.8 − 2.0) = 390 N

(c) Three flat (horizontal) pieces: 590 N from 0 to 2 s, then a jump down to 490 N from 2 to 8 s, then a jump down to 390 N from 8 to 10 s. The pieces are flat because each acceleration is constant.

(d) N = m(g + a). In the diagram at t = 1 s, a is positive, so N must be bigger than mg: the up arrow is longer. On the graph the first piece (590 N) sits above the weight line (490 N), the middle piece equals the weight (a = 0, the arrows would be equal), and the last piece sits below it (a negative, the up arrow would be shorter). All three representations say the same thing.

(e) Displacement is the area under the v-t graph: ½(2)(4.0) + (6)(4.0) + ½(2)(4.0) = 4 + 24 + 4 = 32 m.

Scoring (10 points)

• (a) 2 points: 1 point for exactly two correctly labeled forces with correct directions; 1 point for N longer than mg.
• (b) 3 points: 1 point for each correct reading (590 N, 490 N, 390 N). An acceleration found from the slope must be shown at least once.
• (c) 2 points: 1 point for flat pieces with jumps (no slopes or curves); 1 point for the correct order of levels (highest first, lowest last), consistent with (b).
• (d) 2 points: 1 point for N = m(g + a); 1 point for a sentence linking at least two representations (diagram arrow lengths and graph levels) to the sign of a.
• (e) 1 point for 32 m with work shown (area or kinematics).

(a) Clamp the spring to the stand so it hangs straight down. Measure the position of the spring's bottom end with the metre stick with no mass on it. Hang 50 g, wait until it stops bouncing, and measure the new position. The stretch is the difference. Repeat for 100 g, 150 g, 200 g and 250 g. Read the metre stick at eye level, and repeat each reading (for example remove and re-hang the masses) and average.

(b) Vertical: the weight F = mg of the hanging mass (equal to the spring force at rest). Horizontal: the stretch Δx. Since F = kΔx, the graph is a straight line through the origin and its slope is k.

(c) Weights: 0.49, 0.98, 1.47, 1.96, 2.45 N. Graph weight versus stretch. A best-fit line through the points has slope about (2.45 − 0.49) N / (0.098 − 0.020) m = 1.96 / 0.078 = 25 N/m. (Use two points on your best-fit line, not just data points, if the line misses them.) k ≈ 25 N/m.

(d) Same slope, so the same k: full length = natural length + stretch, so each length is shifted by the same constant. Shifting every point sideways by the same amount does not change the slope. The line would not pass through the origin; it would cross the length axis at the natural length (where the force is zero).

(e) Weight = 0.40 · 9.8 = 3.92 N. Δx = F/k = 3.92/25 = 0.16 m. Condition: the spring must still obey Hooke's law at this force (it is not stretched past its elastic limit). This is outside the tested range, so it is an extrapolation.

Scoring (10 points)

• (a) 3 points: 1 point for measuring stretch as a change from the unloaded position; 1 point for using several different masses; 1 point for a valid way to reduce error (repeat trials, eye-level reading, wait for it to stop).
• (b) 2 points: 1 point for force (or weight) versus stretch, or stretch versus force; 1 point for the correct link to k (slope = k, or slope = 1/k if axes are swapped).
• (c) 2 points: 1 point for converting mass to weight; 1 point for k between 24 and 26 N/m from a slope.
• (d) 2 points: 1 point for "same k" with the reason that each point shifts by the same constant; 1 point for "does not pass through the origin" (intercept at natural length).
• (e) 1 point for about 0.16 m with a valid condition (stays in the linear, elastic range).

(a) At the top, the center of the circle is below the car. To move in a circle, the car needs a net force toward the center, which is down. Gravity pulls down and the road pushes up, so the weight must be bigger than the normal force. The faster the car goes, the bigger the net downward force it needs to curve that tightly. Gravity does not change, so the road must push up less: the normal force gets smaller as speed increases. At some speed the road pushes with zero force. Above that speed, gravity alone is not enough to curve the car along the road, so the car leaves the road surface (it "gets air").

(b) Take down (toward the center) as positive: mg − N = mv²/R, so

N = m(g − v²/R)

Set N = 0: g = v²/R, so v = √(gR).

(c) In N = m(g − v²/R), the term v²/R grows as v grows, so N gets smaller: this matches "the road pushes less." N reaches zero exactly when v²/R = g, which is the speed √(gR): this matches "at some speed the road pushes with zero force." For v larger than √(gR), the equation would give a negative N, which a road cannot provide (it can only push), so the car leaves the road: this matches the last claim.

(d) v²/R = 144/40 = 3.6 m/s². N = 1200(9.8 − 3.6) = 1200 · 6.2 = 7440 N ≈ 7400 N (less than the weight, 11 760 N). Loss of contact: v = √(9.8 · 40) = √392 = 19.8 m/s.

(e) The same. v = √(gR) has no mass in it. The truck needs twice the net force to curve, but its weight is also twice as big, so the two effects cancel.

Scoring (10 points)

• (a) 3 points: 1 point for saying the net force points down, toward the center; 1 point for "normal force decreases as speed increases" with the reason that a larger net force is needed and gravity is constant; 1 point for describing the car losing contact at high speed. The paragraph must be logical and free of contradictions (for example no outward "centrifugal force").
• (b) 2 points: 1 point for mg − N = mv²/R (or the same with consistent signs); 1 point for v = √(gR).
• (c) 2 points: 1 point for linking the growing v²/R term to smaller N; 1 point for linking N = 0 (or negative N) to losing contact.
• (d) 2 points: 1 point for N ≈ 7400 N; 1 point for 19.8 m/s.
• (e) 1 point for "the same" with mass cancelling as the reason.