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Questions on Newton's laws and the resultant force
Newton's laws say that a resultant force changes motion rather than sustaining it, and that only the resultant force enters F = ma. A free-body diagram drawn before any algebra is what makes that resultant possible to find. A third-law pair is one type of force shared between two bodies, which is why weight and the normal contact force are not a pair.
19 original questions · 60 marks · the Newton's laws and the resultant force notes · Mechanics
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State Newton's first law of motion.
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An object stays at rest, or continues at constant velocity (1), unless acted on by a resultant external force (1).A mass of 2.0 kg has an acceleration of 3.0 m s−2. Calculate the resultant force acting on it.
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F = ma = 2.0 × 3.0 (1)
F = 6.0 N (1)Explain what is meant by the resultant force on an object.
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The resultant force is the single force equivalent to all the individual forces acting on the object (1); it is their vector sum (1).A delivery van travels along a straight, level road at a constant velocity of 20 m s−1. State the magnitude of the resultant force acting on it.
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Zero: at constant velocity there is no resultant force (Newton's first law). (1)A model rocket has a mass of 250 g. At one instant during its flight its acceleration is 12 m s−2. Calculate the resultant force on the rocket at this instant.
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Mass = 0.250 kg (1). F = ma = 0.250 × 12 = 3.0 N. (1)State Newton's third law of motion.
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When body A exerts a force on body B, body B exerts a force on body A that is equal in magnitude and opposite in direction (1); the two forces are of the same type and act on different bodies. (1)A block of mass 4.0 kg is pushed to the right by a horizontal force of 12 N while a frictional force of 4.0 N acts to the left. Calculate the resultant force on the block and its acceleration.
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Resultant = 12 − 4.0 = 8.0 N to the right (1)
a = F/m = 8.0/4.0 (1)
a = 2.0 m s−2 (1)A person of mass 70 kg stands in a lift. Taking g = 9.81 m s−2, calculate (a) their weight and (b) the force the lift floor exerts on them when the lift accelerates upwards at 2.0 m s−2.
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(a) W = mg = 70 × 9.81 (1)
W = 687 N (1)
(b) N − mg = ma, so N = m(g + a) = 70 × (9.81 + 2.0) (1)
N = 827 N (1)A 3.0 kg block and a 2.0 kg block are in contact on a smooth surface. A horizontal force of 10 N pushes the 3.0 kg block, which pushes the 2.0 kg block. Calculate (a) the acceleration of the blocks and (b) the contact force between them.
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(a) a = F/(total mass) = 10/5.0 (1)
a = 2.0 m s−2 (1)
(b) Contact force = 2.0 × 2.0 (1)
Contact force = 4.0 N (1)A book rests on a table. Identify the Newton's third-law pair to the force of the table pushing up on the book, and explain why the book's weight is not that pair.
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The third-law pair is the book pushing down on the table with an equal and opposite force (1). The weight is not the pair because it is a gravitational force from the Earth (1), a different type of force acting on the same object (1).A crane raises a 350 kg load on a vertical cable. The tension in the cable is 3.65 kN. Taking g = 9.81 m s−2, calculate the acceleration of the load.
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Resultant force = T − mg = 3650 − 350 × 9.81 = 216.5 N upwards (1). a = F/m (1) = 216.5/350 = 0.62 m s−2 upwards. (1)An ice hockey puck of mass 160 g slides in a straight line across level ice. Its speed falls from 18 m s−1 to 12 m s−1 in 4.0 s. Calculate the average frictional force acting on the puck.
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a = Δv/Δt = (18 − 12)/4.0 = 1.5 m s−2 (1). F = ma = 0.160 × 1.5 (1) = 0.24 N, directed opposite to the motion. (1)Explain, using Newton's third law, how a swimmer is able to accelerate forwards through the water.
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The swimmer's hands and feet push backwards on the water (1). By Newton's third law the water exerts an equal and opposite force forwards on the swimmer (1). This forward force gives a resultant force on the swimmer, so they accelerate (F = ma). (1)A 2.0 kg block slides down a smooth slope inclined at 30° to the horizontal. Taking g = 9.81 m s−2, calculate its acceleration down the slope.
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Component of weight down the slope = mg sinθ (1)
a = g sin 30° = 9.81 × 0.5 (1)
a = 4.9 m s−2 (2 s.f.) (1)A 3.0 kg mass hangs from a light string that passes over a smooth pulley and connects to a 2.0 kg mass on a smooth horizontal table. Taking g = 9.81 m s−2, calculate (a) the acceleration of the system and (b) the tension in the string.
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(a) For the system: a = m1g/(m1 + m2) (1)
a = 3.0 × 9.81/5.0 (1)
a = 5.89 m s−2 (1)
(b) For the table mass: T = m2a = 2.0 × 5.89 (1)
T = 11.8 N (1)A person stands at rest on horizontal ground. State the Newton's third-law pair to the force the person exerts on the ground, and explain why the person remains in equilibrium.
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The person pushes down on the ground; the third-law pair is the ground pushing up on the person with an equal and opposite (normal contact) force (1). That upward force and the person's weight both act on the person, so they are not a third-law pair (1). They are equal in magnitude, so the resultant force on the person is zero and they remain in equilibrium (1).A packing crate of mass 12 kg rests on a ramp inclined at 25° to the horizontal. The maximum frictional force the ramp can exert on the crate is 55 N. Taking g = 9.81 m s−2, deduce whether the crate remains at rest on the ramp.
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Component of weight down the slope = mg sin θ (1) = 12 × 9.81 × sin 25° = 49.8 N (1). This is less than the 55 N maximum frictional force available (1), so friction can balance it: the resultant force is zero and the crate remains at rest. (1)A student of mass 62.0 kg stands on scales, calibrated in kilograms, on the floor of a lift. While the lift is moving, the scales read 56.5 kg. Taking g = 9.81 m s−2, determine the magnitude and direction of the lift's acceleration, and state one possible motion of the lift at this moment.
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Force from the scales N = 56.5 × 9.81 = 554 N (1). Newton's second law (up positive): N − mg = ma (1), so a = (554.3 − 62.0 × 9.81)/62.0 = −0.87, i.e. 0.87 m s−2 (1) directed downwards (1). The lift could be moving upwards and slowing down, or moving downwards and speeding up: the reading fixes the acceleration, not the velocity. (1)A student says: “When a horse pulls a cart, Newton's third law means the cart pulls back on the horse just as hard, so the forces cancel and the pair can never accelerate.” Explain fully why the student's conclusion is wrong. Refer to the forces acting on the cart, the forces acting on the horse, and Newton's laws.
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The pull on the cart and the pull on the horse act on different bodies, so they can never cancel: each body's motion depends only on the forces acting on that body (1). Forces on the cart: the forward pull from the horse and backward resistive forces (1). The cart accelerates when the horse's pull exceeds the resistance, giving a resultant force (F = ma) (1). The horse's hooves push backwards on the ground (1); by the third law the ground pushes forwards on the horse, and when this exceeds the backward pull of the cart plus resistance the horse accelerates too (1). So both accelerate together: third-law pairs act on different bodies and never cancel within one body's free-body diagram. (1)
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