8.1 Mechanics: Motion, Newton's Laws, Work, and Energy

8.1 Mechanics: Motion, Newton's Laws, Work, and Energy

Mechanics is the branch of physics that AdUCET science items draw on most heavily, because almost every question can be built from five or six formulas and finished with mental arithmetic. Adamson University publishes no content outline for AdUCET, so treat the topics below as standard Philippine Grade 11-12 General Physics 1 coverage rather than an official blueprint. Waves, heat, electricity and magnetism are handled separately in section 8.2.

Scalars, Vectors, and the Vocabulary of Motion

A scalar is fully described by a magnitude and a unit: 5 kg, 12 s, 40 m. A vector needs a magnitude and a direction: 40 m east, 9.8 m/s$^2$ downward. Most motion traps come from swapping one for the other.

Worked orientation example. A student jogs 300 m north along Taft Avenue, turns around, and jogs 100 m south, taking 200 s in total.

1. Distance $= 300 + 100 = 400$ m; displacement $= 300 - 100 = 200$ m north.
2. Average speed $= 400 \div 200 = 2$ m/s.
3. Average velocity $= 200 \div 200 = 1$ m/s north.

Same trip, two different numbers. Acceleration is the rate of change of velocity, $a = (v_f - v_i)/t$, in m/s$^2$; an object accelerates when it speeds up, slows down, or merely changes direction.

The four kinematic equations

For uniform acceleration in a straight line, memorise which variable each equation leaves out and choose by elimination.

For free fall, replace $a$ with $g = 9.8$ m/s$^2$ (round to $10$ for calculator-free work) and set $v_i = 0$ for a dropped object. Mass is irrelevant: a one-peso coin and a bowling ball fall together when air resistance is negligible.

Worked example 1 — free fall from a building

A stone is dropped from the roof of a 45 m building. Using $g = 10$ m/s$^2$:

• Step 1. No final speed is given, so use $d = \tfrac{1}{2}gt^2$: $45 = \tfrac{1}{2}(10)t^2 = 5t^2$.
• Step 2. $t^2 = 9$, so $t = 3$ s.
• Step 3. $v_f = gt = 10 \times 3 = 30$ m/s.
• Check. $v_f^2 = 2gd = 2(10)(45) = 900$, and $\sqrt{900} = 30$ m/s. The two routes agree.

Reading motion graphs

• Distance-time graph: the slope is speed. A horizontal line means at rest, a steeper line means faster, and a curving line means the speed is changing.
• Velocity-time graph: the slope is acceleration, and the area under the line is the displacement. A line below the axis means motion in the reverse direction.

Worked example 2 — distance from a velocity-time graph

A tricycle accelerates uniformly from rest to 12 m/s in 4 s, cruises at 12 m/s for 6 s, then brakes uniformly to rest in 2 s. Break the area into three shapes:

1. Triangle: $\tfrac{1}{2}(4)(12) = 24$ m
2. Rectangle: $6 \times 12 = 72$ m
3. Triangle: $\tfrac{1}{2}(2)(12) = 12$ m

Total displacement $= 24 + 72 + 12 = 108$ m.

Projectile motion

A projectile's horizontal and vertical motions are independent. Horizontal velocity stays constant (ignoring air resistance) while the vertical motion is ordinary free fall. Two consequences appear constantly in tests: a ball rolled off a table and a ball simply dropped from the same height hit the floor at the same time, and the flight path is a parabola whose range is greatest at a launch angle of $45^\circ$.

Newton's Three Laws of Motion

1. First law (inertia). A body keeps its state of rest or of uniform straight-line motion unless a net external force acts. Inertia is measured by mass, which is why a loaded truck is harder to stop than a bicycle.
2. Second law. $F_{net} = ma$. One newton is the force that gives 1 kg an acceleration of 1 m/s$^2$. Acceleration always points in the direction of the net force, not of the largest single force.
3. Third law. If A pushes B, then B pushes A with equal magnitude in the opposite direction. The pair acts on two different bodies, so the two forces can never cancel on one free-body diagram.

The classic trap reads: if every force has an equal and opposite partner, how can anything accelerate? The answer is that the partner force is applied to the other object. When you push a cart, your push acts on the cart and the cart's push acts on you.

Mass, weight, and the normal force

Weight is a force: $F_g = mg$. A 60 kg student weighs $60 \times 9.8 = 588$ N on Earth, $60 \times 1.6 = 96$ N on the Moon, and $60 \times 3.7 = 222$ N on Mars, yet the mass stays 60 kg everywhere. The normal force ($N$) is the surface's perpendicular push back; on level ground with nothing else pressing down, $N = mg$.

Friction

Static friction holds a stationary object and grows to match whatever push you apply, up to a maximum $f_s = \mu_s N$. Kinetic friction, $f_k = \mu_k N$, is roughly constant once sliding starts, and $\mu_s > \mu_k$ — which is exactly why a heavy aparador takes a hard shove to start moving and a gentler push to keep moving.

        Free-body diagram: crate pushed across a floor
                       N (normal, up)
                       ^
                       |
      f_k <------- [ CRATE ] -------> F_applied
                       |
                       v
                      mg (weight, down)

   Vertical: N = mg  ->  no vertical acceleration
   Horizontal: F_net = F_applied - f_k  ->  a = F_net / m

Momentum, Impulse, and Collisions

Momentum is $p = mv$, in kg$\cdot$m/s, and it is a vector. Impulse is $F\Delta t = \Delta p$: spreading a collision over a longer contact time reduces the force, which is the whole design principle behind helmet padding, car crumple zones, and pulling your hands back as you catch a ball.

In a closed system, total momentum before a collision equals total momentum after. A 1,200 kg car moving at 15 m/s rear-ends a stationary 800 kg car and the two lock together: $p = 1{,}200 \times 15 = 18{,}000$ kg$\cdot$m/s, total mass $= 2{,}000$ kg, so the wreck moves off at $18{,}000 \div 2{,}000 = 9$ m/s.

Work, Power, and Mechanical Energy

Work is $W = Fd\cos\theta$, in joules. Work is zero in three situations: no displacement (straining against a concrete wall), force perpendicular to displacement ($\cos 90^\circ = 0$, as when you carry a bag horizontally), or no force at all. Power is $P = W/t$, in watts; 1 horsepower is about 746 W.

• Kinetic energy: $KE = \tfrac{1}{2}mv^2$. Because speed is squared, doubling speed quadruples the kinetic energy — the reason stopping distances balloon.
• Gravitational potential energy: $PE = mgh$, measured from whatever reference height you choose.
• Conservation of mechanical energy: when only gravity does work, $KE_i + PE_i = KE_f + PE_f$.

Worked example 3 — power on a staircase

A 50 kg student runs up stairs 4 m high in 8 s, with $g = 10$ m/s$^2$.

1. Work against gravity: $W = mgh = 50 \times 10 \times 4 = 2{,}000$ J.
2. Power: $P = W/t = 2{,}000 \div 8 = 250$ W.

Worked example 4 — energy conservation on a falling mango

A 3 kg mango drops from a branch 5 m up, with $g = 10$ m/s$^2$.

1. At the top: $PE = 3 \times 10 \times 5 = 150$ J and $KE = 0$.
2. Just before impact: $PE = 0$, so $KE = 150$ J.
3. Solve for speed: $\tfrac{1}{2}(3)v^2 = 150 \Rightarrow v^2 = 100 \Rightarrow v = 10$ m/s.
4. Halfway down (2.5 m): $PE = 75$ J and $KE = 75$ J — the total is still 150 J.

Simple Machines and Mechanical Advantage

Mechanical advantage is $MA = \text{load force} \div \text{effort force}$; the ideal mechanical advantage is $IMA = \text{effort distance} \div \text{load distance}$. A machine multiplies force, never energy: ideally work in equals work out, and real machines lose part of the input to friction, so efficiency $= (W_{out}/W_{in}) \times 100%$ is always below 100%.

• Lever: $MA = $ effort arm $\div$ load arm. First class has the fulcrum in the middle (seesaw, scissors), second class has the load in the middle (wheelbarrow, nutcracker), third class has the effort in the middle (tongs, fishing rod, the human forearm) and trades force for speed with $MA < 1$.
• Pulley: a fixed pulley has $MA = 1$ and only redirects the effort; a movable pulley gives $MA = 2$; a block and tackle gives $MA$ equal to the number of rope strands supporting the load.
• Inclined plane: $IMA = \text{slope length} \div \text{height}$. A 6 m ramp rising 1.5 m has $IMA = 6 \div 1.5 = 4$, so a 200 N crate ideally needs only $200 \div 4 = 50$ N of push. Energy still balances: $200 \times 1.5 = 300$ J lifted, $50 \times 6 = 300$ J pushed.

What Examiners Test

1. Distance versus displacement in a there-and-back trip — the classic zero-displacement item.
2. Whether you read slope or area off a graph. Area under a velocity-time line is displacement; slope of a distance-time line is speed.
3. Mass unchanged, weight changed, when an object is moved to the Moon or Mars.
4. Third-law pairs, testing whether you know the two forces act on different bodies.
5. Zero-work situations, especially carrying a load horizontally at constant speed.
6. The squared term in $KE = \tfrac{1}{2}mv^2$ — tripling the speed multiplies kinetic energy by nine, not by three.