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**Physics Support Materials Higher Mechanics and Properties of Matter**

Solutions to Problems Newton’s 2nd law, energy and power Click on a question number 52, 53, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75

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**Physics Support Materials Higher Mechanics and Properties of Matter**

Newton’s 2nd law, energy and power 52 A lift of mass 500 kg travels upwards at a constant speed. Calculate the tension in the lifting cable: The weight of the lift is 500 x 9.8 = 4900 N The lift is travelling at constant speed so the forces are balanced. Therefore the tension is 4900 N Click the mouse to continue

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**Physics Support Materials Higher Mechanics and Properties of Matter**

Newton’s 2nd law, energy and power 53 A fully loaded tanker has a mass of 2.0 x 108 kg. As the speed of the tanker increases from zero to a steady maximum speed of 8.0 ms-1 the force from the propellers remains constant at 3.0 x 106 N. a) i) Calculate the acceleration of the tanker just as it starts from rest. ii) What is the size of the force of friction acting on the tanker when it is travelling at the steady speed of 8.0 ms-1 ? The motion is a constant speed so the forces are balanced. The friction force is therefore 3.0 x 106 N. b) When its engines are stopped, the tanker takes 50 minutes to come to rest from a speed of 8.0 ms-1.Calculate its average deceleration. Click the mouse to continue

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**Physics Support Materials Higher Mechanics and Properties of Matter**

Newton’s 2nd law, energy and power 55 Two girls push a car of mass 2000 kg. Each applies a force of 50 N and the force of friction is 60 N. Calculate the acceleration of the car. The resultant force is (2 x 50) - 60 = 40 N. Click the mouse to continue

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**Physics Support Materials Higher Mechanics and Properties of Matter**

Newton’s 2nd law, energy and power 56 A boy on a skateboard rides up a slope. The total mass of the boy and the skateboard is 90 kg. He decelerates from 12 ms-1 to 2 ms-1 in 6 seconds. Calculate the resultant force acting on him. Click the mouse to continue

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**Physics Support Materials Higher Mechanics and Properties of Matter**

Newton’s 2nd law, energy and power 57 A box is pulled along a rough surface with a constant force of 140 N. If the mass of the box is 30 kg and it accelerates at 4 ms-2 calculate: a) the unbalanced force causing the acceleration. b) the force of friction between the box and the surface. The friction force is 140 N N = 20 N. Click the mouse to continue

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**Physics Support Materials Higher Mechanics and Properties of Matter**

Newton’s 2nd law, energy and power 58 An 800 kg Metro is accelerated from 0 to 18 ms-1 in 12 s. a) What is the resultant force acting on the Metro? b) How far does the car travel in these 12 s? At the end of the 12 s period the brakes are operated and the car comes to rest in a distance of 50 m. c) What is the average frictional force acting on the car? Click the mouse to continue

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**Physics Support Materials Higher Mechanics and Properties of Matter**

Newton’s 2nd law, energy and power 59 A rocket of mass kg is launched vertically upwards. Its engines produce a constant thrust of N. a) i) Draw a diagram showing all the forces acting on the rocket. Thrust = N ii) Calculate the initial acceleration of the rocket. The resultant force is = N. b) As the rocket rises its acceleration is found to increase. Give three reasons for this. i) The fuel is burnt up so the mass decreases. ii) The air is thinner so the friction is reduced. Weight = mg = x 9.8 = N iii) The gravitational field strength is less so the weight is reduced. Click the mouse to continue

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**Physics Support Materials Higher Mechanics and Properties of Matter**

Newton’s 2nd law, energy and power 59 continued A rocket of mass kg is launched vertically upwards. Its engines produce a constant thrust of N. c) Calculate the acceleration of the same rocket from the surface of the Moon if the Moon’s gravitational field strength is 1.6 N kg-1. Thrust = N The resultant force is = N. d) Explain in terms of Newton’s laws of motion why a rocket can travel from the Earth to the Moon and for most of the journey not burn up any fuel . In outer space there is no friction. According to N1, on object with no forces on it will continue at constant speed. The only time fuel is needed is when it is required to speed up or slow down (N2). Weight = mg = x 1.6 =64000 N Click the mouse to continue

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**Physics Support Materials Higher Mechanics and Properties of Matter**

Newton’s 2nd law, energy and power 60 A rocket takes off and accelerates to 90 ms-1 in 4 s. The resultant force acting on it is 40 kN upwards. a) Calculate the mass of the rocket. Engine force Weight Friction b) Calculate the force produced by the rocket’s engines if the average force of friction is 5000 N. The resultant force = engine force - force of friction - weight 40000 = engine force (1778 x 9.8) engine force = engine force = N Click the mouse to continue

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**Physics Support Materials Higher Mechanics and Properties of Matter**

Newton’s 2nd law, energy and power 61 What is the minimum force required to lift a helicopter of mass 2000 kg upwards with an initial acceleration of 4 ms-2 ? Air resistance is 1000 N. The lifting force = force to hover + force to overcome friction + force to accelerate The lifting force = mg ma The lifting force = x x 4 The lifting force = The lifting force = N Click the mouse to continue

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**Physics Support Materials Higher Mechanics and Properties of Matter**

Newton’s 2nd law, energy and power 62 A crate of mass 200 kg is placed on a balance on a lift. a) What would be the reading on the balance, in newtons, when the lift was stationary? The reading is the weight = mg = 200 x = N b) The lift now accelerates upwards at 1.5 m s-2. What is the new reading on the balance? The reading is the weight (mg) + force required to accelerate lift (ma) The reading = x = N c) The lift now travels up at a constant speed of 5 m s-1. What is the reading on the balance? The speed is constant so the forces are balanced. The reading is the weight = 1960 N d) For the last stage of the journey calculate the reading on the balance when the lift decelerates at 1.5 m s-2 while moving up. The reading is the weight (mg) - force required to decelerate lift (ma) The reading = x = N Click the mouse to continue

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**Physics Support Materials Higher Mechanics and Properties of Matter**

Newton’s 2nd law, energy and power 63 A small lift in a hotel is fully loaded and has a mass of 250 kg. For safety reasons the tension in the pulling cable must never be greater than 3500 N. a)What is the tension in the cable when the lift is: i) at rest The tension is the weight of the lift = mg = 250 x 9.8 = N ii) moving up at a constant 1 m s-1 The tension is the weight of the lift = mg = N iii) accelerating upwards at 2 m s-2 The tension = weight (mg) + accelerating force (ma) The tension = x = N iv) accelerating downwards at 2 m s-2? The tension = weight (mg) - accelerating force (ma) The tension = x = N Click the mouse to continue

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**Physics Support Materials Higher Mechanics and Properties of Matter**

Newton’s 2nd law, energy and power 63 continued A small lift in a hotel is fully loaded and has a mass of 250 kg. For safety reasons the tension in the pulling cable must never be greater than 3500 N. b) Calculate the maximum permitted upward acceleration of the lift. The maximum tension = weight (mg) + maximum accelerating force (ma) c) Describe a situation where the lift could have an upward acceleration greater than the value in b) without breaching safety regulations. If the lift was not fully loaded it could have a higher acceleration. Click the mouse to continue

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**Physics Support Materials Higher Mechanics and Properties of Matter**

Newton’s 2nd law, energy and power 64 A package of mass 4 kg is hung from a spring balance attached to the ceiling of a lift which is accelerating at 3 ms-2 . What is the reading on the spring balance? 3 m s-2 4 kg Reading on balance = weight (mg) + accelerating force (ma) Reading on balance = 4 x x = N Click the mouse to continue

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**Physics Support Materials Higher Mechanics and Properties of Matter**

Newton’s 2nd law, energy and power 65 The graph shows how the downward speed of a lift varies with time. a / ms-2 4 t / s 0.5 1 -1 v / ms-1 4 t / s 2 a) Draw the corresponding acceleration / time graph. b) A 4 kg mass is suspended from a spring balance inside the lift. Determine the reading on the balance at each stage of the motion. For the first 4 s the reading on balance = m(g-a) = 4 ( ) = N From the 4th to the 10th second speed is constant so the reading on balance = mg = 4 x 9.8 = N For the last 2 s the reading on balance = m(g+a) = 4 ( ) = N Click the mouse to continue

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**Physics Support Materials Higher Mechanics and Properties of Matter**

Newton’s 2nd law, energy and power 66 Two masses are pulled along a flat surface as shown below. 2 kg 1 kg 24 N T Find a) the acceleration of the masses. b) the tension, T. The tension, T, is the accelerating force on the 2 kg mass Click the mouse to continue

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**Physics Support Materials Higher Mechanics and Properties of Matter**

Newton’s 2nd law, energy and power 67 A car of mass 1200 kg tows a caravan of mass 1000 kg. The frictional forces acting on the car and the caravan are 200 N and 500 N respectively. The car accelerates at 2 ms-2 . a) Calculate the force exerted by the engine on the car. b) What force does the tow bar exert on the caravan? The resultant force acting on the caravan is F The resultant force = force exerted by tow bar - frictional force Force exerted by tow bar = = N Click the mouse to continue

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**Physics Support Materials Higher Mechanics and Properties of Matter**

Newton’s 2nd law, energy and power 67 continued c) The car then travels at a constant speed of 10 ms-1. Assuming the frictional forces to be unchanged, calculate the new engine force and the force exerted by the towbar on the caravan. The speed of the car / caravan is constant so the forces are balanced. The engine force needs only to equal the friction force. This totals 700 N. The force exerted by the towbar is equal to the friction force on the caravan. This equals 500 N. d) The car brakes and decelerates at 5 ms-2. Calculate the force exerted by the brakes. (Assume the frictional forces remain constant.) The resultant force acting on the car / caravan is F Note: all forces act in the opposite direction to the motion The resultant force = force exerted by brakes + frictional force Force exerted by brakes= = N Click the mouse to continue

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**Physics Support Materials Higher Mechanics and Properties of Matter**

Newton’s 2nd law, energy and power 68 A tractor of mass 1200 kg pulls a log of mass 400 kg. The tension in the tow rope is 2000 N and the frictional force on the log is 800 N. How far will the log move in 4 s assuming it was stationary to begin with? 2000 N Friction 800 N The resultant force acting on the log is 1200 N. Click the mouse to continue

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**Physics Support Materials Higher Mechanics and Properties of Matter**

Newton’s 2nd law, energy and power 69 A force of 60 N pushes three blocks as shown. 60 N C B A If each block has a mass of 8 kg and the force of friction on each block is 4 N calculate: a) the acceleration of the blocks The resultant force acting on the blocks is (3 x 4) = N. b) the force block A exerts on block B Find the resultant force, F, acting on blocks B and C. Resultant force, F = force exerted by A on B - friction forces. Click the mouse to continue Force exerted by A on B = ( 2 x 4 ) = 40 N

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**Physics Support Materials Higher Mechanics and Properties of Matter**

Newton’s 2nd law, energy and power 69 60 N C B A Calculate: c) the force block B exerts on block C Find the resultant force, F, acting on block C Resultant force, F = force exerted by B on C - friction force. Force exerted by B on C = = 20 N The pushing force is then reduced until the blocks move at a constant speed. d) Calculate the value of this pushing force The motion is a constant speed, so the forces are balanced. The force required is equal to the frictional forces which is 12 N. e) Does the force block A exerts on block B now equal the force block B exerts on block C? No. A exerts 8 N on B and C to overcome friction. B exerts 4 N on C to overcome friction. Click the mouse to continue

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**Physics Support Materials Higher Mechanics and Properties of Matter**

Newton’s 2nd law, energy and power 70 A trolley is connected by string to a 1 kg mass as shown. The bench and pulley are frictionless. 2 kg 1 kg a) Calculate the acceleration of the trolley. b) Calculate the tension in the string. The tension in the string is the force that accelerates the 2 kg trolley. Click the mouse to continue

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**Physics Support Materials Higher Mechanics and Properties of Matter**

Newton’s 2nd law, energy and power 71 A force of 800 N is applied to a canal barge by means of a rope angled at 40o to the direction of the canal. If the mass of the barge is 1000 kg and the force of friction between the barge and the water is 100 N, find the acceleration of the barge. 800 N 40o 1000 kg Friction100 N Click the mouse to continue

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**Physics Support Materials Higher Mechanics and Properties of Matter**

Newton’s 2nd law, energy and power 72 A crate of mass 100 kg is pulled along a rough surface at the angles shown. 100 kg 120 N 20o a) If the crate is moving at a constant speed of 1 ms-1 what is the force of friction? The crate is moving at constant speed so the forces are balanced. The friction force is N. b) If the forces were increased to 140 N at the same angle, calculate the acceleration of the crate? Assume the friction force remains at N. The resultant force is therefore = N Click the mouse to continue

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**Physics Support Materials Higher Mechanics and Properties of Matter**

Newton’s 2nd law, energy and power 73 A 2 kg block of wood is placed on the slope shown. It remains stationary. What is the size of the frictional force acting up the slope? 30o m g cos 60 60o m g The block of wood is stationary so the forces are balanced. The force of friction up the slope is 9.8 N. Click the mouse to continue

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**Physics Support Materials Higher Mechanics and Properties of Matter**

Newton’s 2nd law, energy and power 74 A 500 g trolley runs down a runway which is 2 m long and raised 30 cm at one end. If its speed remains constant throughout, calculate the force of friction acting up the slope. 2m 0.3 m m g cos 81.37 81.37o qo m g The trolley has a constant speed so the forces are balanced. The force of friction up the slope is N. Click the mouse to continue

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**Physics Support Materials Higher Mechanics and Properties of Matter**

Newton’s 2nd law, energy and power 75 The brakes on a car fail while it is parked at the top of a hill. It runs down the hill for a distance of 50 m until it crashes into a hedge. The mass of the car is 900 kg and the hill makes an angle of 15o to the horizontal. If the force of friction is 300 N find: 15 o Friction 300 N 50 m a) the component of the weight acting down the slope m g cos 75 75 o m g b) the acceleration of the car The resultant force is = N Click the mouse to continue

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**Physics Support Materials Higher Mechanics and Properties of Matter**

Newton’s 2nd law, energy and power 75 The brakes on a car fail while it is parked at the top of a hill. It runs down the hill for a distance of 50 m until it crashes into a hedge. The mass of the car is 900 kg and the hill makes an angle of 15o to the horizontal. If the force of friction is 300 N find: 15 o Friction 300 N 50 m m g sin 75 c) the speed of the car as it hits the hedge m g cos 75 75 o m g d) the force acting perpendicular to the car (the reaction) when it is on the hill Click the mouse to continue

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