Connect force, mass and motion on the same diagram.

About 44.1 metres, or 145 feet, ignoring air resistance — from h = ½ g t² with g = 9.80665 m/s². It arrives at 29.4 m/s, roughly 66 mph. The acceleration is what makes this counter-intuitive: it covers 4.9 m in the first second and 24.5 m in the third.

No. In the absence of air, everything accelerates at the same 9.80665 m/s² regardless of mass, because a heavier object needs proportionally more force to accelerate it and gravity supplies exactly that. In real air, shape and density decide: a feather loses to a hammer because of drag, not gravity.

Because 9.80665 m/s² is the internationally agreed standard value of gravitational acceleration, fixed by the CGPM in 1901, and every other unit built on force — the pound-force, the kilogram-force, psi, the foot-pound — is defined from it. Using 9.81 introduces a 0.034% error that quietly breaks the round trip between derived units. The actual local value where you are is a different question again, varying about ±0.3% with latitude and altitude.

Because the work needed to accelerate an object is force times distance, and at higher speed you cover more distance during each interval of acceleration. Integrating that gives ½ m v². The practical consequence is severe: a 40% speed increase doubles the energy that has to be absorbed in a crash.

Energy is a stored quantity, measured in joules. Work is the transfer of energy by a force acting through a distance, also in joules. Power is the rate at which that transfer happens, in watts — joules per second. Lifting the same box the same height is the same work whether you take one second or one minute; only the power differs.

Yes, and it is the standard classroom example. If the force is perpendicular to the motion, cos 90° = 0 and the work is exactly zero: carrying a suitcase along a level corridor does no work on the suitcase. Your muscles still burn energy holding it, but that energy goes into your body, not into the load.

Multiply the coefficient of static friction by the normal force. On level ground the normal force is the weight, m g. A 100 kg crate on a floor with μ = 0.5 needs about 490 N — roughly 110 pounds-force — just to break loose, and slightly less to keep it moving once it is going.

It is the stiffness: how many newtons of force each metre of deflection produces, so N/m. Automotive and industrial specs often use lb/in instead, and 1 lb/in equals 175.1268 N/m. A soft laboratory spring might be 20 N/m; a car coil spring is tens of thousands.

Angular momentum L = I ω is conserved when nothing outside applies a torque. Pulling the arms in cuts the moment of inertia I, so ω has to rise to keep L the same. Notice that rotational energy ½ I ω² does not stay constant — it goes up, and the increase comes from the muscular work of pulling the arms against the outward pull.

No. The period depends only on length and gravity: T = 2π √(L/g). Mass cancels out, because a heavier bob experiences proportionally more restoring force. This is precisely why pendulums made good clocks and good gravimeters — the one thing hardest to control, the mass, does not enter the answer.

A full there-and-back cycle takes 2.006 seconds at standard gravity, so almost exactly one second in each direction. To get a period of exactly two seconds — the classic seconds pendulum used in clock design — the length has to be 0.9940 m, about 39.14 inches.

Compare its average density with the fluid’s. If the object’s mass divided by its total volume is less than the fluid density, it floats, and it settles at the depth where the displaced fluid weighs exactly what the object does. Steel ships float because the hull encloses mostly air, so the ship’s average density is well below that of water.

Seawater is about 1,025 kg/m³ against fresh water’s 1,000, so each cubic metre displaced lifts about 2.5% harder. A vessel therefore needs 2.5% less submerged volume to support the same weight and rides visibly higher — the reason load lines on a hull are marked separately for fresh and salt water.