Forces, balanced & unbalanced

Year 7 answers

Fluency

A push or pull, measured in newtons (N).
Contact: friction, normal, tension, applied, air resistance. Non-contact: gravity, magnetic, electrostatic.
The single force that has the same effect as all the individual forces on an object combined.
Mass is how much matter is in an object (kg) — it does not change. Weight is the gravitational force on that mass (N) — it depends on gravity.
$25 - 10 = 15$ N to the right.
$m = W/g = 200/10 = 20$ kg.
A diagram showing every force acting on an object as an arrow from a single point, with correct direction and relative size.
Net force $= 50 - 50 = 0$ N. The rope does not accelerate; it stays at rest (or moves at constant speed if already moving).
It continues at constant speed in a straight line forever (no friction, no other forces).
$W = 60 \times 10 = 600$ N.

Reasoning

Gravity pulls the book down; the table pushes it up with an equal normal force. The two forces cancel, giving zero net force — so the book’s motion does not change.
Balanced. Constant velocity in a straight line means zero net force; the driving force equals air resistance + rolling friction.
Forces: gravity (weight) downward, air resistance equal and upward. Net force $= 0$ , so the skydiver falls at constant (terminal) speed.
From $F = ma$ , for the same acceleration, a larger mass requires a proportionally larger force.
Mass is the amount of matter — unchanged by location. Gravity on the Moon is about $1/6$ of Earth’s, so the gravitational force (weight) is $1/6$ — but the amount of matter is identical.
No. Although speed is constant, direction is changing (circular motion). A change in direction means the velocity changes, which requires a net force — gravity, pulling toward Earth.

Reasoning

$F_{\text{net}} = 5000 - 4000 = 1000$ N upward. The rocket accelerates upward.
$a = F/m = 12/3 = 4$ m/s $^2$ .
(a) On jumping: gravity much greater than air resistance → large net force down → accelerates down. (b) As speed grows, air resistance grows → net downward force shrinks → acceleration decreases. (c) At terminal velocity: air resistance = weight → net force zero → constant speed.
The post is pulled at an angle between north and east. The pull is stronger toward north (80 N > 60 N), so the direction is closer to north than to east — roughly north-north-east.

Reasoning

$a = \Delta v/\Delta t = 20/10 = 2$ m/s $^2$ . $F = ma = 1200 \times 2 = 2400$ N. This force is generated by friction between the drive wheels and the road (the engine spins the wheels; the ground pushes the car forward).
When a lift accelerates upward, the floor must push harder than gravity to both support the person and accelerate them upward — so the normal force is greater than weight, making you feel heavier. Going down: floor pushes less than gravity briefly, so you feel lighter.
Mass: $W/g = 50/10 = 5$ kg, unchanged. On the new planet: $W = 5 \times 30 = 150$ N. The astronaut feels three times heavier — walking, lifting and standing all become very tiring.
By Newton’s third law the forces are equal and opposite. By $F = ma$ the skater with less mass ( $40$ kg) has greater acceleration and moves faster ( $60$ kg skater moves slower). Ratio of speeds: $60:40 = 3:2$ .