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Forces and Newton's laws questions
Draw the forces, add them up, divide by the mass. Newton's second law is three instructions long, and almost every mechanics mark hangs off the first one, the diagram.
6 original questions · 21 marks · the forces and newton's laws notes · Mechanics
Every question here is written for this library rather than taken from a past paper. Write your answer out before opening the worked one: the answers award marks point by point, and the marks are easier to see when you have something of your own to compare against.
State Newton's first law, and what it implies about the forces on a car cruising at a steady 30 m s⁻¹ in a straight line.
Worked answer
A body keeps a constant velocity unless a resultant force acts on it. Steady speed in a straight line therefore means zero resultant, so the driving force exactly balances the resistances. The car's speed is irrelevant to that conclusion. Fast and steady needs no more resultant force than slow and steady. B1 for the law, B1 for the zero resultant.A 1000 kg car accelerates at 2 m s⁻² against a total resistance of 500 N. Find the driving force.
Worked answer
F = ma applies to the resultant, so D − 500 = 1000 × 2 and D = 2500 N. The engine has two jobs, causing the acceleration and paying off the resistance. Forgetting the second gives 2000 N and loses the accuracy mark. M1 for the equation of motion, A1 for 2500 N.A 6 kg crate is dragged along a floor by a horizontal rope of tension 42 N against friction of 12 N. Find the acceleration, and the normal reaction from the floor.
Worked answer
Along the motion the resultant is 42 − 12 = 30 N, so a = 30/6 = 5 m s⁻². Vertically the crate neither rises nor falls, so R = mg = 6 × 9.8 = 58.8 N. The two directions are handled separately, and the vertical equation is an equilibrium even though the crate is accelerating horizontally. M1 for the horizontal resultant, A1 for a = 5, M1 for R = mg, A1 for 58.8 N. Draw the force diagram before writing either equation; the four forces on it are the first mark.A person of mass 70 kg stands on scales in a lift. Taking g = 9.8 m s⁻², find the scale reading (the normal reaction) when the lift accelerates upwards at 1.5 m s⁻², and when it accelerates downwards at 1.5 m s⁻².
Worked answer
Take upwards as positive. Accelerating up, R − mg = ma gives R = 70(9.8 + 1.5) = 791 N, heavier than the true weight of 686 N. Accelerating down, mg − R = ma gives R = 70(9.8 − 1.5) = 581 N, lighter. The scales read the normal reaction, not the weight, and the reaction is whatever F = ma demands. Both answers must straddle 686 N; if they do not, a sign has gone wrong. M1 for R − mg = ma, A1 for 791 N, M1 for mg − R = ma, A1 for 581 N.A book rests on a table. Identify the Newton's third law partner of the normal reaction on the book, and explain why the reaction and the book's weight are not a third-law pair.
Worked answer
The partner is the push of the book down on the table. Same type of force, same magnitude, opposite direction, acting on the other body. The weight's own partner is the gravitational pull of the book on the Earth. Reaction and weight act on the same body and can differ in size, as they do in a lift, so they fail both tests for a third-law pair. B1 for naming the partner, B1 for the weight's own partner, B1 for the reason the pairing fails.A car of mass 900 kg travelling at 20 m s⁻¹ brakes uniformly to rest in 50 m on a level road. Find the braking force. The same braking force is then applied at 20 m s⁻¹ while the car descends a slope inclined at 5° to the horizontal. Find the new stopping distance.
Worked answer
On the level, v2 = u2 + 2as gives 0 = 400 + 100a, so a = −4 m s⁻² and the braking force is F = 900 × 4 = 3600 N against the motion. On the slope the weight now has a component down the hill of 900 × 9.8 × sin 5° = 769 N, which works with the car. Taking down the slope as positive, 769 − 3600 = 900a gives a = −3.15 m s⁻². Then 0 = 400 + 2(−3.15)s, so s = 63.6 m. M1 A1 for the suvat step and the braking force, M1 for resolving the weight along the slope, A1 for the new acceleration, M1 A1 for the stopping distance. Two ideas are being tested at once. Suvat supplies the acceleration and Newton's second law prices it in force. On the slope, only the component mg sin θ acts along the motion, and the fact that the car takes 13.6 m longer to stop is the whole point of the question.
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