Newton's Laws
Write the equation connecting net force, mass, and acceleration. What are the SI units of each?
ΣF = ma, a vector relationship. Net force: N = kg·m/s²; mass: kg; acceleration: m/s². Use net force, not an arbitrary single force.
Write the weight equation. Find the weight of a 3.0 kg object.
W = mg = (3.0)(9.8) ≈ 29 N downward. Mass and weight have different units.
Write the equations for average velocity, average speed, and average acceleration. Explain what goes in each numerator.
v_avg = Δx/Δt; average speed = total distance/Δt; a_avg = (v_f − v_i)/Δt. Displacement is signed in one dimension; distance is path length; acceleration uses change in velocity.
| During ideal projectile motion, what are a_x and a_y?
Write the horizontal displacement equation.
a_x = 0; a_y = −g.
Δx = v_0x t.
Horizontal velocity remains constant in ideal flight.
A student writes Δx = vt² for motion at constant velocity. Use units to identify the error and correct the equation.
(m/s)s² = m·s, not m. At constant velocity, Δx = vt; its units are meters.
A cart moves right at constant velocity. What are its acceleration and net force? Can individual forces still act?
a = 0 and ΣF = 0. Individual forces may be nonzero and balanced. Constant velocity includes constant direction.
A force F acts at angle θ above horizontal. Write its horizontal and vertical components.
F_x = F cos θ; F_y = F sin θ for a force directed up/right with angle measured from +x. Other quadrants require appropriate signs.
| Fill the missing terms:
v_f = v_i + ____;
Δx = v_i t + ____;
v_f² = v_i² + ____.
State the shared assumption.
at;
½at²;
2aΔx.
All three require constant acceleration over the interval.
A ball launches at 20 m/s at 30° above horizontal. Write the component equations and find v_0x and v_0y.
v_0x = v_0 cos θ = 20 cos 30° ≈ 17.3 m/s;
v_0y = v_0 sin θ = 10 m/s.
Accept sensible rounding.
A student says, “Negative acceleration always means slowing down.” Give a signed velocity/acceleration example that disproves the claim.
Example: v = −4 m/s and a = −2 m/s². The object moves left and speeds up while both signs remain negative. Speed changes according to the relationship between velocity and acceleration, not the acceleration sign alone.
A 4.0 kg box experiences 18 N right and 6.0 N left. Write a signed force equation and find acceleration.
Right positive: 18 − 6.0 = 4.0a_x, so a_x = +3.0 m/s².
A 5.0 kg box is pulled by 20 N at 30° above horizontal on a level floor. It has no vertical acceleration. Draw its vertical forces, write their balance, and find the normal force.
N upward, mg downward, and the pull's vertical component upward. N + 20 sin 30° − 5.0(9.8) = 0. N = 49 − 10 = **39 N**. The upward pull reduces the required support force.
A cart starts from rest with constant acceleration +3.0 m/s² for 4.0 s. Find final velocity and displacement.
v_f = 0 + (3.0)(4.0) = +12 m/s.
Δx = 0 + ½(3.0)(4.0)² = +24 m.
Write the equations for vertical velocity and vertical displacement with upward positive. At the highest point of an angled projectile, which velocity component is zero? Is acceleration zero?
v_y = v_0y − gt; Δy = v_0y t − ½gt².
At the peak, v_y = 0; v_x remains nonzero for an angled launch with a horizontal component. Acceleration remains downward at 9.8 m/s².
A walkway moves right at 1.5 m/s relative to the ground. A person walks left at 0.50 m/s relative to the walkway. Write the relative-velocity relationship and find the person's ground velocity.
v_person/ground = v_person/walkway + v_walkway/ground = −0.50 + 1.5 = **+1.0 m/s**, right.
Frame labels prevent mixing velocities measured by different observers.
Earth pulls downward on a backpack. Name the third-law partner. Are the backpack's weight and the floor's normal force a third-law pair? Explain.
The backpack pulls gravitationally on Earth toward the backpack. Weight and normal force act on the same backpack, so they are not third-law partners. Interaction partners act on different objects.
| A box remains stationary on a 30° incline. Write the magnitude of the component of weight along the incline, the normal force, and the static friction required. There are no other forces besides gravity and contact forces.
Weight along slope: mg sin 30°, downhill. N = mg cos 30°. Actual static friction f_s = mg sin 30°, uphill, provided the surface can supply this force. Do not assume f_s equals its maximum value.
A car travels at +10 m/s and brakes at constant acceleration −2.0 m/s² until it stops. Choose an equation that avoids time and find braking displacement.
v_f² = v_i² + 2aΔx. Δx = (0 − 100)/(−4.0) =+25 m.
Negative acceleration can accompany positive displacement.
A ball leaves a 1.25 m-high table horizontally at 3.0 m/s. Its flight time is given as 0.505 s. Find horizontal displacement and vertical velocity just before impact.
Δx = (3.0)(0.505) = 1.515 m, about 1.5 m.
v_y = 0 − (9.8)(0.505) = −4.949 m/s, about −4.9 m/s.
Negative indicates downward.
A cart starts from rest and accelerates uniformly to 8.0 m/s in 4.0 s. A student uses Δx = v_f t and gets 32 m. Explain the mistake and calculate the correct displacement.
Final velocity is not the velocity throughout the interval. For constant acceleration, Δx = ½(v_i + v_f)t = ½(0 + 8.0)(4.0) = 16 m.
The same net force acts on two carts. Cart B has three times the mass of cart A. Compare their accelerations using an equation. If the force on B then triples, how does its acceleration compare with A's original acceleration?
a = Fnet/m
Initially a_B = a_A/3
Tripling B's net force gives a_B
New = 3F_net/(3m_A) = a_A
A 0.20 kg object hangs at the midpoint of a symmetric line. Each side makes 30° above horizontal. Write the vertical equilibrium equation and find the tension in each side. What happens to tension if the angle decreases?
2T sin 30° − mg = 0. T = (0.20)(9.8)/(2 sin 30°) = **1.96 N**, about 2.0 N. A smaller angle requires greater tension. At zero angle, finite tension cannot supply the needed upward component in this ideal model.
A car travels at +12 m/s during a 0.50 s reaction interval, then brakes at constant −3.0 m/s² until it stops. Write the equation for each interval and find total stopping distance.
Reaction: Δx_r = v_i t_r = (12)(0.50) = 6.0 m. Braking: Δx_b = (v_f² − v_i²)/(2a) = (0 − 144)/(−6.0) = 24 m.
Total = 30 m.
Separate intervals because their accelerations differ.
A ball leaves a 4.9 m-high platform horizontally at 6.0 m/s. Derive the flight time from vertical motion, then find horizontal displacement. If horizontal launch speed doubles, what happens to flight time and horizontal displacement?
v_0y = 0 and Δy = −4.9 m: −4.9 = −½(9.8)t².
Positive solution t = 1.0 s.
Δx = (6.0)(1.0) = 6.0 m.
Doubling v_0x leaves time unchanged and doubles displacement to 12 m.
This depends on the same height and initial vertical velocity, with air resistance ignored.
A student says, “The reaction force cancels the action force, so a projectile has zero net force after release.” Explain the error, name the force acting during ideal flight, and write the net-force equations in x and y.
Third-law forces act on different objects and cannot cancel in the projectile's own force sum. After release, only gravity acts in the ideal model. ΣF_x = 0 = ma_x; ΣF_y = −mg = ma_y, so a_y = −g. Its gravitational partner is the projectile's pull on Earth.