physics support materials higher mechanics and properties of matter b solutions to problems -...

Physics Support MaterialsPhysics Support MaterialsHigher Higher Mechanics and Properties of MatterMechanics and Properties of Matter

Solutions to Problems - Newton’s 2nd law, energy and powerSolutions to Problems - Newton’s 2nd law, energy and power

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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A lift of mass 500 kg travels upwards at a constant speed. Calculate the A lift of mass 500 kg travels upwards at a constant speed. Calculate the tension in the lifting cable:tension in the lifting cable:

52 Newton’s 2nd law, energy and power

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The weight of the lift is 500 x 9.8 = 4900 The weight of the lift is 500 x 9.8 = 4900 NNThe lift is travelling at constant speed so The lift is travelling at constant speed so the forces are balanced.the forces are balanced.

Therefore the tension is 4900 NTherefore the tension is 4900 N


A fully loaded tanker has a mass of 2.0 x 10A fully loaded tanker has a mass of 2.0 x 1088 kg. As the speed of the tanker increases from zero to kg. As the speed of the tanker increases from zero to a steady maximum speed of 8.0 msa steady maximum speed of 8.0 ms-1-1 the force from the propellers remains constant at 3.0 x 10 the force from the propellers remains constant at 3.0 x 1066 N. N.

53


Newton’s 2nd law, energy and power

a) i) Calculate the acceleration of the tanker just as it starts from rest.a) i) Calculate the acceleration of the tanker just as it starts from rest.

mF

a 8

6

102103

a 2-s m 015.0a

ii) What is the size of the force of friction acting on the tanker when it is 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 mstravelling at the steady speed of 8.0 ms-1-1 ??

The motion is a constant speed so the forces are balanced. The friction force is The motion is a constant speed so the forces are balanced. The friction force is therefore 3.0 x 10therefore 3.0 x 1066 N. N.

b) When its engines are stopped, the tanker takes 50 minutes to come to rest from b) When its engines are stopped, the tanker takes 50 minutes to come to rest from a speed of 8.0 msa speed of 8.0 ms-1-1.Calculate its average deceleration..Calculate its average deceleration.

tuv

a

605080

a 2-s m 0027.0

30008 a


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 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.car.

55



The resultant force is (2 x 50) - 60 = 40 N.The resultant force is (2 x 50) - 60 = 40 N.

mF

a 2-s m 02.02000

40 a


A boy on a skateboard rides up a slope. The total mass of the boy and the skateboard is 90 kg. 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 msHe decelerates from 12 ms-1-1 to 2 ms to 2 ms-1-1 in 6 seconds. Calculate the resultant force acting on him. in 6 seconds. Calculate the resultant force acting on him.

56



tuv

a 2-s m 67.1

6122 a

maF N 150)67.1(90 F


A box is pulled along a rough surface with a constant force of 140 N. If the 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 msmass of the box is 30 kg and it accelerates at 4 ms-2-2 calculate: calculate:

57



a) the unbalanced force causing the acceleration.a) the unbalanced force causing the acceleration.

maF N 120430 Fb) the force of friction between the box and the surface.b) the force of friction between the box and the surface.

The friction force is 140 N - 120 N = 20 N.The friction force is 140 N - 120 N = 20 N.


An 800 kg Metro is accelerated from 0 to 18 msAn 800 kg Metro is accelerated from 0 to 18 ms-1-1 in 12 s. in 12 s.

58



a) What is the resultant force acting on the Metro?a) What is the resultant force acting on the Metro?

tuv

a 2-s m 5.1

12018 a

maF N 12005.1800 F

b) How far does the car travel in these 12 s?b) How far does the car travel in these 12 s?

2

21atuts 2125.1

21

120 s m 108sAt the end of the 12 s period the brakes are operated and the car comes to rest in a At the end of the 12 s period the brakes are operated and the car comes to rest in a distance of 50 m. distance of 50 m.

c) What is the average frictional force acting on the car?c) What is the average frictional force acting on the car?

asuv 222 502180 22 a 2-s m 24.3100

324 a

maF N 2592

)24.3(800

F

F


A rocket of mass 40000 kg is launched vertically upwards. Its engines produce a A rocket of mass 40000 kg is launched vertically upwards. Its engines produce a constant thrust of 700000 N.constant thrust of 700000 N.

59



a) i) Draw a diagram showing all the forces acting on the rocket.a) i) Draw a diagram showing all the forces acting on the rocket.

Weight = mg = 40000 x 9.8

=392000 N

Thrust = 700000 N

ii) Calculate the initial acceleration of the rocket.ii) Calculate the initial acceleration of the rocket.

mF

a 2-s m 7.740000

308000 a

The resultant force is 700000 - 392000 = 308000 N.The resultant force is 700000 - 392000 = 308000 N.

b) As the rocket rises its acceleration is found to b) As the rocket rises its acceleration is found to increase. Give three reasons for this.increase. Give three reasons for this.

i) The fuel is burnt up so the mass decreases.i) The fuel is burnt up so the mass decreases.

ii) The air is thinner so the friction is reduced.ii) The air is thinner so the friction is reduced.

iii) The gravitational field strength is less so the iii) The gravitational field strength is less so the weight is reduced.weight is reduced.


A rocket of mass 40000 kg is launched vertically upwards. Its engines produce a A rocket of mass 40000 kg is launched vertically upwards. Its engines produce a constant thrust of 700000 N.constant thrust of 700000 N.

59 continued



Weight = mg = 40000 x 1.6

=64000 N

Thrust = 700000 N

c) Calculate the acceleration of the same rocket from c) Calculate the acceleration of the same rocket from the surface of the Moon if the Moon’s gravitational the surface of the Moon if the Moon’s gravitational field strength is 1.6 N kgfield strength is 1.6 N kg-1-1..

mF

a 2-s m 9.1540000

636000 a

The resultant force is 700000 - 64000 = 636000 N.The resultant force is 700000 - 64000 = 636000 N.

d) Explain in terms of Newton’s laws of motion why a d) Explain in terms of Newton’s laws of motion why a rocket can travel from the Earth to the Moon and for rocket can travel from the Earth to the Moon and for most of the journey not burn up any fuel .most of the journey not burn up any fuel .

In outer space there is no friction. According to N1, In outer space there is no friction. According to N1, on object with no forces on it will continue at on object with no forces on it will continue at constant speed. The only time fuel is needed is constant speed. The only time fuel is needed is when it is required to speed up or slow down (N2).when it is required to speed up or slow down (N2).


A rocket takes off and accelerates to 90 msA rocket takes off and accelerates to 90 ms-1-1 in 4 s. The resultant force in 4 s. The resultant force acting on it is 40 kN upwards.acting on it is 40 kN upwards.

60



a) Calculate the mass of the rocket.a) Calculate the mass of the rocket.

tuv

a 2-s m 5.22

4090 a

maF aF

m kg 17785.22

40000 m

b) Calculate the force produced by the rocket’s engines if b) Calculate the force produced by the rocket’s engines if the average force of friction is 5000 N.the average force of friction is 5000 N.

The resultant force = engine force - force of friction - weightThe resultant force = engine force - force of friction - weight

40000 = engine force - 5000 - (1778 x 9.8)40000 = engine force - 5000 - (1778 x 9.8)

Engine force

Weight

Frictionengine force =40000 + 5000 + 17424engine force =40000 + 5000 + 17424

engine force = 62424 Nengine force = 62424 N


What is the minimum force required to lift a helicopter of mass 2000 kg What is the minimum force required to lift a helicopter of mass 2000 kg upwards with an initial acceleration of 4 msupwards with an initial acceleration of 4 ms-2 -2 ? Air resistance is 1000 N.? Air resistance is 1000 N.

61



The lifting force = force to hover + force to overcome friction + force to accelerate The lifting force = force to hover + force to overcome friction + force to accelerate

The lifting force = mg + 1000 + maThe lifting force = mg + 1000 + ma

The lifting force = 2000 x 9.8 + 1000 + 2000 x 4The lifting force = 2000 x 9.8 + 1000 + 2000 x 4

The lifting force = 19600 + 1000 + 8000The lifting force = 19600 + 1000 + 8000

The lifting force = 28600 NThe lifting force = 28600 N


62


A crate of mass 200 kg is placed on a balance on a lift.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?a) What would be the reading on the balance, in newtons, when the lift was stationary?

The reading is the weight = mgThe reading is the weight = mg = 200 x 9.8 = 1960 N= 200 x 9.8 = 1960 N

b) The lift now accelerates upwards at 1.5 m sb) The lift now accelerates upwards at 1.5 m s-2-2. What is the new reading on the balance?. What is the new reading on the balance?

The reading is the weight (mg) + force required to accelerate lift (ma)The reading is the weight (mg) + force required to accelerate lift (ma)

The reading = 1960 + 200 x 1.5 = 2260 NThe reading = 1960 + 200 x 1.5 = 2260 N

c) The lift now travels up at a constant speed of 5 m sc) The lift now travels up at a constant speed of 5 m s -1-1. What is the reading on the balance?. What is the reading on the balance?

The speed is constant so the forces are balanced. The reading is the weight = 1960 The speed is constant so the forces are balanced. The reading is the weight = 1960 NNd) For the last stage of the journey calculate the reading on the balance when the lift decelerates d) For the last stage of the journey calculate the reading on the balance when the lift decelerates

at 1.5 m sat 1.5 m s-2-2 while moving up. while moving up.

The reading is the weight (mg) - force required to decelerate lift (ma)The reading is the weight (mg) - force required to decelerate lift (ma)

The reading = 1960 - 200 x 1.5 = 1660 NThe reading = 1960 - 200 x 1.5 = 1660 N


A small lift in a hotel is fully loaded and has a mass of 250 kg. For safety 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.reasons the tension in the pulling cable must never be greater than 3500 N.


63

a)What is the tension in the cable when the lift is:a)What is the tension in the cable when the lift is:

i) at resti) at rest


The tension is the weight of the lift = mg = 2450 NThe tension is the weight of the lift = mg = 2450 N

ii) moving up at a constant 1 m sii) moving up at a constant 1 m s-1-1

iii) accelerating upwards at 2 m siii) accelerating upwards at 2 m s-2-2

The tension = weight (mg) + accelerating force (ma)The tension = weight (mg) + accelerating force (ma)

The tension is the weight of the lift = mgThe tension is the weight of the lift = mg = 250 x 9.8 = 2450 N= 250 x 9.8 = 2450 N

The tension = 2450 + 250 x 2 = 2950 NThe tension = 2450 + 250 x 2 = 2950 N

iv) accelerating downwards at 2 m siv) accelerating downwards at 2 m s-2-2??

The tension = weight (mg) - accelerating force (ma)The tension = weight (mg) - accelerating force (ma)

The tension = 2450 - 250 x 2 = 1950 NThe tension = 2450 - 250 x 2 = 1950 N


A small lift in a hotel is fully loaded and has a mass of 250 kg. For safety 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.reasons the tension in the pulling cable must never be greater than 3500 N.


63 continued Newton’s 2nd law, energy and power

b) Calculate the maximum permitted upward acceleration of the lift.b) Calculate the maximum permitted upward acceleration of the lift.

The maximum tension = weight (mg) + maximum accelerating force (ma)The maximum tension = weight (mg) + maximum accelerating force (ma)

a 250245035002-s m 2.4

25024503500 a

c) Describe a situation where the lift could have an upward acceleration greater c) Describe a situation where the lift could have an upward acceleration greater than the value in b) without breaching safety regulations.than the value in b) without breaching safety regulations.

If the lift was not fully loaded it could have a higher acceleration.If the lift was not fully loaded it could have a higher acceleration.


A package of mass 4 kg is hung from a spring balance attached to the ceiling of a lift A package of mass 4 kg is hung from a spring balance attached to the ceiling of a lift which is accelerating at 3 mswhich is accelerating at 3 ms-2 -2 . What is the reading on the spring balance?. What is the reading on the spring balance?

64



3 m s-2

4 kg

Reading on balance = weight (mg) + accelerating force (ma)Reading on balance = weight (mg) + accelerating force (ma)

Reading on balance = 4 x 9.8 + 4 x 3 = 51.2 NReading on balance = 4 x 9.8 + 4 x 3 = 51.2 N


The graph shows how the downward speed of a lift varies with time.The graph shows how the downward speed of a lift varies with time.

65



a) Draw the corresponding acceleration / time graph.a) Draw the corresponding acceleration / time graph.

v / ms-1

4 10 12 t / s

2

a / ms-2

4 10 12 t / s

0.5

1

-1

b) A 4 kg mass is suspended from a spring balance inside the lift. Determine 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.the reading on the balance at each stage of the motion.

For the first 4 s the reading on balance = m(g-a) = 4 ( 9.8 - 0.5 ) = 37.2 NFor the first 4 s the reading on balance = m(g-a) = 4 ( 9.8 - 0.5 ) = 37.2 N

From the 4th to the 10th second speed is constant so From the 4th to the 10th second speed is constant so the reading on balance = mg = 4 x 9.8 = 39.2 Nthe reading on balance = mg = 4 x 9.8 = 39.2 N

For the last 2 s the reading on balance = m(g+a) = 4 ( 9.8 + 1 ) = 43.2 NFor the last 2 s the reading on balance = m(g+a) = 4 ( 9.8 + 1 ) = 43.2 N


Two masses are pulled along a flat surface as shown below.Two masses are pulled along a flat surface as shown below.

66



Find a) the acceleration of the masses.Find a) the acceleration of the masses.

2 kg 1 kg 24 NT

mF

a 2-s m 8324 a

b) the tension, T.b) the tension, T.

maF N 1682 TFThe tension, T, is the accelerating force on the 2 kg massThe tension, T, is the accelerating force on the 2 kg mass


A car of mass 1200 kg tows a caravan of mass 1000 kg. The frictional forces acting on the car 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 msand the caravan are 200 N and 500 N respectively. The car accelerates at 2 ms-2-2 . .

67



a) Calculate the force exerted by the engine on the car.a) Calculate the force exerted by the engine on the car.

maF force,Resultant N 440022200 Fforces frictional - force engine forceResultant

700forceengine4400 N 51007004400forceengine

b) What force does the tow bar exert on the caravan?b) What force does the tow bar exert on the caravan?

maF N 200021000 FThe resultant force acting on the caravan is FThe resultant force acting on the caravan is F

The resultant force = force exerted by tow bar - frictional forceThe resultant force = force exerted by tow bar - frictional force

Force exerted by tow bar = 2000 + 500 = 2500 NForce exerted by tow bar = 2000 + 500 = 2500 N


67 continued



c) The car then travels at a constant speed of 10 msc) The car then travels at a constant speed of 10 ms -1-1. Assuming the frictional . Assuming the frictional forces to be unchanged, calculate the new engine force and the force exerted by the forces to be unchanged, calculate the new engine force and the force exerted by the towbar on the caravan.towbar on the caravan.

The speed of the car / caravan is constant so the forces are balanced. 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 The engine force needs only to equal the friction force. This totals 700 N. N. The force exerted by the towbar is equal to the friction force on the The force exerted by the towbar is equal to the friction force on the caravan. This equals 500 N.caravan. This equals 500 N.

d) The car brakes and decelerates at 5 msd) The car brakes and decelerates at 5 ms-2-2. Calculate the force exerted by the . Calculate the force exerted by the brakes. (Assume the frictional forces remain constant.)brakes. (Assume the frictional forces remain constant.)

maF N 1100052200 F

The resultant force acting on the car / caravan is FThe resultant force acting on the car / caravan is F

The resultant force = force exerted by brakes + frictional forceThe resultant force = force exerted by brakes + frictional force

Force exerted by brakes= 11000 - 700 = 10300 NForce exerted by brakes= 11000 - 700 = 10300 N

Note: all forces act in the opposite direction to the Note: all forces act in the opposite direction to the motionmotion


A tractor of mass 1200 kg pulls a log of mass 400 kg. The tension in the tow rope is 2000 N and the 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 frictional force on the log is 800 N. How far will the log move in 4 s assuming it was stationary to begin with?begin with?

68



2000 N Friction

800 N

The resultant force acting on the log is 1200 N.The resultant force acting on the log is 1200 N.

mF

a 2-s m 3400

1200 a

2

21atuts 243

21

40 s

m 24s


A force of 60 N pushes three blocks as shown.A force of 60 N pushes three blocks as shown.

69



If each block has a mass of 8 kg and the force of friction on each block is 4 N calculate: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 blocksa) the acceleration of the blocks

60 N C B A

mF

a 2-s m 22448 a

The resultant force acting on the blocks is 60 - (3 x 4) = 48 N.The resultant force acting on the blocks is 60 - (3 x 4) = 48 N.

b) the force block A exerts on block Bb) the force block A exerts on block B Find the resultant force, F, acting on Find the resultant force, F, acting on blocks B and C.blocks B and C.maF N 32216 F

Resultant force, F = force exerted by A on B - friction forces.Resultant force, F = force exerted by A on B - friction forces.

Force exerted by A on B = 32 + ( 2 x 4 ) = 40 NForce exerted by A on B = 32 + ( 2 x 4 ) = 40 N


69



Calculate:Calculate:c) the force block B exerts on block Cc) the force block B exerts on block C

60 N C B A

Find the resultant force, F, acting on block CFind the resultant force, F, acting on block C

maF N 1628 FResultant force, F = force exerted by B on C - friction force.Resultant force, F = force exerted by B on C - friction force.

Force exerted by B on C = 16 + 4 = 20 NForce exerted by B on C = 16 + 4 = 20 N

The pushing force is then reduced until the blocks move at a constant speed.The pushing force is then reduced until the blocks move at a constant speed.d) Calculate the value of this pushing forced) Calculate the value of this pushing forceThe motion is a constant speed, so the forces are balanced. The force required is The motion is a constant speed, so the forces are balanced. The force required is equal to the frictional forces which is 12 N.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 e) Does the force block A exerts on block B now equal the force block B exerts on block C?block C?No. A exerts 8 N on B and C to overcome friction. B exerts 4 N on C to No. A exerts 8 N on B and C to overcome friction. B exerts 4 N on C to overcome friction.overcome friction.


A trolley is connected by string to a 1 kg mass as shown. The bench and A trolley is connected by string to a 1 kg mass as shown. The bench and pulley are frictionless.pulley are frictionless.

70



a) Calculate the acceleration of the trolley.a) Calculate the acceleration of the trolley.

2 kg

1 kg

mF

a 2-s m 27.338.9 a

b) Calculate the tension in the string.b) Calculate the tension in the string.

maF N 54.627.32 F

The tension in the string is the force that accelerates the 2 kg trolley.The tension in the string is the force that accelerates the 2 kg trolley.


A force of 800 N is applied to a canal barge by means of a rope angled at 40A force of 800 N is applied to a canal barge by means of a rope angled at 40oo to to the direction of the canal. If the mass of the barge is 1000 kg and the force of 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 friction between the barge and the water is 100 N, find the acceleration of the barge. barge.

71



800 N

40o

1000 kgFriction100 N

40cos800 force pulling ofcomponent Horizontal N 8.612N 512.8 1008.612forceResultant

mF

a 2-s m 513.01000

8.512 a


A crate of mass 100 kg is pulled along a rough surface at the angles A crate of mass 100 kg is pulled along a rough surface at the angles shown.shown.

72



a) If the crate is moving at a constant speed a) If the crate is moving at a constant speed of 1 msof 1 ms-1 -1 what is the force of friction?what is the force of friction? 100 kg

120 N

120 N

20o

20o

)20cos120(2 forces pulling ofcomponent Horizontal N 5.225The crate is moving at constant speed so the forces are balanced. The crate is moving at constant speed so the forces are balanced. The friction force is 225.5 N.The friction force is 225.5 N.

b) If the forces were increased to 140 N at the same angle, calculate the b) If the forces were increased to 140 N at the same angle, calculate the acceleration of the crate?acceleration of the crate?

)20cos140(2 forces pulling ofcomponent Horizontal N 1.263Assume the friction force remains at 225.5 N. The resultant force is Assume the friction force remains at 225.5 N. The resultant force is therefore 263.1 - 225.5 = 37.6 Ntherefore 263.1 - 225.5 = 37.6 N

mF

a 2-s m 376.0100

6.37 a


A 2 kg block of wood is placed on the slope shown. It remains stationary. 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?What is the size of the frictional force acting up the slope?

73



30o m g

60om g cos 60

N8.95.08.9260cos mgThe block of wood is stationary so the forces are balanced. The The block of wood is stationary so the forces are balanced. The force of friction up the slope is 9.8 N.force of friction up the slope is 9.8 N.


A 500 g trolley runs down a runway which is 2 m long and raised 30 cm at one end. If 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.its speed remains constant throughout, calculate the force of friction acting up the slope.

74



m g

81.37om g cos 81.37

2m 0.3 m

o

23.0

sin o63.8

15.08.95.037.81cos mgN 735.0

The trolley has a constant speed so the forces are balanced. The The trolley has a constant speed so the forces are balanced. The force of friction up the slope is 0.735 N.force of friction up the slope is 0.735 N.


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 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 15m until it crashes into a hedge. The mass of the car is 900 kg and the hill makes an angle of 15oo to the to the horizontal. If the force of friction is 300 N find: horizontal. If the force of friction is 300 N find:

75



a) the component of the weight acting down the slopea) the component of the weight acting down the slope

15 o

Friction 300 N

50 m

m g

m g cos 7575 o259.08.990075cos mg

N 2283

b) the acceleration of the carb) the acceleration of the car

The resultant force is 2283 - 300 = 1983 N The resultant force is 2283 - 300 = 1983 N

mF

a 2-s m 2.2900

1983 a


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 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 15m until it crashes into a hedge. The mass of the car is 900 kg and the hill makes an angle of 15oo to the to the horizontal. If the force of friction is 300 N find: horizontal. If the force of friction is 300 N find:

75



c) the speed of the car as it hits the hedgec) the speed of the car as it hits the hedge

15 o

Friction 300 N

50 m

m g

m g cos 7575 oasu 2v 22

1-

2

s m 8.14

220

502.220v

v d) the force acting perpendicular to the car (the reaction) when d) the force acting perpendicular to the car (the reaction) when it is on the hillit is on the hill

m g sin 75

966.08.990075sin mgN 8520

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