Scalar definition and examples
A quantity that involves only size
e.g. distance, speed, mass, volume
Vector definition and examples
A quantity involving direction and size
e.g. displacement, velocity, weight
Displacement
Distance and direction from a starting point
(m)
someone walks around a square park with a side length of 40m
determine the :
a) distance walked
b) displacement
a)d=160m
b) s=0m
Someone walks north 100m
Determine the
a) distance
b) displacement
a) d=100m (direction not required)
b)s=100m N (direction required)
Equation for speed
s=d/t
speed=distance/time
Equation for velocity
v=s/t
v=u+at
Convert km/h to m/s
km/h to m/s = divide by 3.6
m/s to km/h = multiply by 3.6
If someone is going 20m/s^-1 W for 12 seconds, calculate the displacement.
s=vt
s=20x12
s=240m W
Calculate and determine how long it will take to drive 80km away if the acceleration is 30ms^-1
t=s/v
t=80000/30
t=2666.67 secs (time is a scalar, so no direction required)
A 50kg crate of tomatoes rests on a 40º incline. If the force of gravity acting on a crate is 500N vertically down towards the ground, calculate the components of the force
a) down the incline
b) perpendicular to the incline
a) down the incline:
=gravity x sin(θ)
=500N x sin(40)
=500 x 0.6528
=3321.4N
b)perpendicular to the incline:
=force of gravity x cos(θ)
500N x cos(40)
= 500N x 0.766
=383N
A 1300kg car is parked on an incline of 30° and then the handbrake is released: (hint: draw it if you can)
Calculate the
a) weight of the car
b) normal force
c) force down the slope
d) acceleration of the car down the incline
a) w=mg
w=1300 x 9.8
w=12740N
b) normal=mgcosθ
normal= 1300 x 9.8 x cos(30)
normal= 12305.9N
c) down the incline= mgsinθ
=1300 x 9.8 x sin(30)
=4297.4N
d) F=ma
3207.4 = 1300 x a
a=2.54ms
Equation for work
w=fs
work= force x displacement
Calculate the quantity of work when pulling a bag of do food 3.5 metres along a table when applying 25N of horizontal force
w=fs
2=25 x 3.5
w=87.5J
Calculate the quantity of work when lifting a 20kg bag of do food at constant speed onto a table 85cm off the ground
w=fs
w=20 x 9.8 x 0.85
w=166.6J
Calculate the amount of work done when: pumping 200L of water at a constant flow rate into a tank 25cm high
w=fs
w=200 x 9.8 x 25
w=49000J
w= 49kj
Equation for kinetic energy
KE= 1/2 x mv2
What is the formula for energy efficiency
efficiency= output/input x 100
Relationship between kinetic energy and gravitational potential energy
As GPE increases, KE decreases and vice versa
When 287J are input in the form of electrical energy, 63J are used to produce light. What is the energy efficiency of the lightbulb.
efficiency= output/input x 100
efficiency= 63/287 x 100
efficiency= 21.95%
A 5kg object falls a certain distance starting from rest and reaching a final velocity at 40m/s. What is the displacement of the object.
mgh=1/2 x mv^2
49h=1/2 x 5 x 40^2
49= 4000
h=81.63m
A 3kg object falls 120m. If it was initially falling at 5m.s, calculate the final velocity of the object.
mgh= 1/2 x mv^2
3 x 9.8 x 120=1/2 x 3 x v^2
3528=1.5v^2
2352=v
v=48.50m/s
Stoichiometry: how many moles of water (H2O) are produced when 2 moles of hydrogen gas (H2) react with oxygen gas (O2)
2 moles of H20
Work and energy: a force of 50N is applied to push a box along a flat surface for 5 metres. How much work is done on the box?
w=fs
w=50 x 5
w= 250J
Vectors and forces: a boat travels 5m east and then 12m north. What is the magnitude of its resultant displacement.
a^2+b^2=c^2
5^2+12^2=c^2
c^2=169
c=13
magnitude is 13 metres
A 2kg watermelon is dropped of a 25m high apartment balcony. If 65J is lost to air resistance, calculate the velocity of the watermelon as it hits the ground.
GPE=KE+heat
mgh=1/2 x mv^2 +65
2 x 9.8 x 25=1/2 x 2v^2 + 65
490=v^2+65
v=20.62m/s
An object with mas m0 is intitially falling at 10m/s. It falls a further distance h, after which it is falling at 25m/s and has lost 20m0J of energy to air resistancne. Calculate the displacement h.
∆KE=KE(final) - KE(initial)
∆KE=1/2 x m0 x 25^2 - 1/2 x m0 x 10^2
∆KE=312.5m0 - 50m0
∆KE=262.5m0
Newton's first law of motion
Newton's first law of motion states that an object will remain at rest or travel at a constant speed in a straight line unless acted upon by an unbalanced force
Newton's second law of motion
Newton's second law of motion states that the net force acting on an object equals the mass of the object multiplied by its acceleration.
(F=ma)
Newton's third law of motion
Newton's third law of motion states that for every action there is an equal and opposite reaction force
Describe using Newton's third law of motion, how a cricket helmet can protect a player's head from the impact force of a cricket ball.
A cricket helmet exerts a reaction force on the on the impact with a cricket ball. This allows the head to not exert the reaction force on impact.
Define temperature
A measure of the average kinetic energy of particles in a material
Define heat
The transfer of energy across the boundary of a system due to a temperature difference
Define conduction
The transfer of heat energy through direct contact
Define convection
The transfer of heat energy through fluid movement
Define radiation
The transfer of heat energy through electromagnetic waves
A 5kg block is attached to a weightless and inflexible rope that has a tension of 70N. What is the acceleration of the block.
F-vectors = T-mg
=70 - 5 x 9.8
= 21N
F-motion= ma
=5a
F-vectors = F-motion
21=5a
a=4.2m/s^-2 up
A 6kg block is suspended by a rope and is accelerating upwards at 2m/s^2. Calculate the tension in the rope.
F-vectors = T - mg
=T - 6 x 9.8
=T - 58.5
F-motion = ma
=6x2
=12
F-vectors = F-motion
T - 58.8=12
T=78.8N
Stoichiometry: Calculate the mass of carbon dioxide (CO2) produced when 44g of propane (C3H8) is burned completely. (molar mass of C3H8=44g/mol, molar mass of CO2=44g/mol)
C3H8 + 5)2 →3CO2 + 4H2)
Mass of CO2=3 mol x 44
g/mol mass of CO2=132g
Work and energy: a 10kg object is lifted to a height of 5m. How much gravitational potential energy does it gain? (g=9.8m/s^2)
mgh=mass x gravity x height
mgh = 10 x 9.8 x 5
mgh=490J
Vectors and forces: a plane flies 300km east and then 400km north. WHat is the magnitude of its resultant displacement?
500km
A hiker stands on top of a 200m cliff. If the hiker's GPE relative to the ground is 156800J, what is the hikers' mass?
mgh= mass x gravity x height
156800 = m x 9.8 x 200
m= 80kg
A diver with a mass of 70kg jumps off a 10m diving platform. Ignoring air resistance, what is the diver's speed just before hitting the water?
mgh= 1/2 x mv^2
70 x 9.8 x 10 = 1/2 x 70v^2
6860 = 30v^2
v^2=196
v=14m/s^-2
A tennis ball is moving at a speed of 25m/s with a kinetic energy of 312.5J. What is the mass of the tennis ball?
mgh=1/2 x mv^2
312.5 = 1/2 x m x 25^2
312.51/2 x m x 625
m=1kg
Work definition and equations
Work is defined as the product of the force and the distance moved in the direction of an applied force.
W=Fs
W=mgh
Energy, work, force relationship
Energy is the ability to do work. Work occurs whenever a force causes movement. A force is a push or pull.
Lever definition
a rigid beam and a fulcrum. The effort (input force) and load (output force) are applied to either end of the beam. The fulcrum is the point on which the beam pivots
Relationship of input distance and required force on a lever
As the input (effort arm) distance increases, less force is needed to lift the mass
Relationship of output distance and required force on a lever
As the output (load arm) distance increases, more force in needed to lift the mass
A person pushes a box with a force of 50N over a distance of 5m along a flat surface. How much work is done on the box?
The work done on the box is 250 J.
A person pulls a sled with a force of 40N at an angle of 30° to the horizontal. If the sled moves 10 meters along the ground, how much work is done?
The work done to pull the sled is 346.4 J
A worker uses a ramp 3m long to raise a 200 kg crate to a height of 1m. Instead of lifting the crate straight up, the worker pushes it along the ramp. How does the length of the ramp affect the amount of effort the worker needs to apply compared to lifting the crate straight up?
The amount of work increases, as the formula W=Fs takes displacement into account, of which the ramp has more of.
An elevator motor raises a 600kg elevator to a height of 10m in 15 seconds. Calculate the work done by the motor and the power output.
Calc work first, then Power = Work / Time
The work done is 58,800 J, and the power output is 3,920 W.
A person uses a lever with an effort arm that is 4m long to lift a rock. The rock is 1m from the pivot point. If the person moves the point where they apply the force closer to the pivot, what do you think happens to the amount of force they need to apply?
The amount of force applied increases, as a shorter input arm is created by moving closer to the fulcrum, thereby increasing the required force.
Final velocity equation
v= u +at
Displacement equation
s= ut + 1/2 at2
Velocity equation with displacement
v2= u2 +2as
Imagine a car is merging onto a highway. The car starts from rest and accelerates uniformly to reach the highway speed limit of 100 km/h (27.8 m/s) in 10 seconds. Calculate the acceleration of the car and the distance it travels during this time.
Given:
Initial velocity u=0 m/s
Final velocity, v=27.8 m/s Time, t=10 s
Step 1: Calculate acceleration (v=u+at )
Step 2: Calculate distance traveled (s= ut + 1/2 at2 ) The car travels 139 meters while accelerating
A person accidentally drops their phone from a balcony that is 15m high. The phone falls under the influence of gravity. Ignoring air resistance, calculate the time it takes for the phone to hit the ground and the speed just before impact.
Given:
Initial velocity u=0 m/s A
cceleration, a=9.8 m/s2
Displacement, s=15 m
Step 1: Calculate time using (s= ut + 1/2 at2 )
Step 2: Calculate the speed just before hitting the ground using (v=u+at )
The speed just before impact is approximately 17.15 m/s
A car is traveling at 30 m/s when the driver applies the brakes. The car decelerates at a constant rate of −6 m/s2 until it comes to a complete stop. How far does the car travel while braking?
Given:
Initial velocity u=30 m/s
Final velocity, v=0 m/s
Acceleration (decelerarion), a=−6 m/s2
Step 1: Use the equation (v2= u2 +2as)
The car travels 75 meters while braking
A rocket is launched from the ground and accelerates uniformly at 20 m/s² for 30 seconds. After 30 seconds, the rocket engines shut off, and it continues upward due to inerƟa. Calculate how high the rocket goes during the powered flight (the first 30 seconds), and determine its final velocity at the end of the powered flight.
Given:
Initial velocity u=0 m/s
Acceleration, a=20 m/s2
Time, t=30 s
Step 1: Calculate the final velocity using (v=u+at ) Step 2: Calculate the height reached during powered flight using (s= ut + 1/2 at2 ) The rocket reaches a height of 9,000 meters during the powered flight phase.
A cyclist accelerates from 5 m/s to 15 m/s in 10 seconds. Calculate the acceleration and the distance covered during this time. Find the cyclist's acceleration and the distance travelled.
Given:
Initial velocity u=5 m/s
Final velocity, v=15 m/s
Time, t=10 s
Step 1: Calculate the acceleration using (v=u+at) Step 2: Calculate the distance using (s= ut + 1/2 at2 ) The cyclist covers 100 meters during this acceleration.
Describe a constant velocity, constant acceleration and constant deceleration on a velocity-time graph
A 2 kg object is moving at 3 m/s. What is its kinetic energy?
𝐾𝐸= 1/2mv2
The object's kinetic energy is 9 J.
A car has a kinetic energy of 25,000 J and a mass of 1,000 kg. What is the velocity of the car?
The car’s velocity is approximately 7.07 m/s.
A 1,200 kg car accelerates from rest to 30 m/s. Calculate the work done to bring the car to this speed.
The work done is 540,000 J
A rock is dropped from a height of 10 meters. If its mass is 5 kg, calculate its speed just before it hits the ground. Ignore air resistance.
GPE=mgh
The rock’s velocity just before hitting the ground is 14 m/s.
A 100 kg bobsled starts from rest at the top of a 30° incline with a height of 20 meters. Assuming no friction, calculate the bobsled’s velocity at the bottom of the incline.
GPE=KE
The bobsled’s velocity at the bottom of the incline is 19.8 m/s.
Conservation of Mechanical Energy Definition
Energy cannot be created or destroyed, only changed from one form to another. In a closed system (with no external forces), the total energy stays the same.
Potential energy definition and equation
Energy of motion
This is the energy an object has due to its position in a gravitational field. The higher the object, the more potential energy it has.
PE=mgh
Kinetic energy definition and equation
The energy an object has because of its motion. The faster an object moves, the more kinetic energy it has
KE= 1/2mv2
Mechanical energy definition and equation
the sum of an object's kinetic energy (energy of motion) and potential energy (energy stored due to position or height)
Etotal= KE + PE
At the top:
The ball has maximum potential energy (PE) and no kinetic energy (KE) if it starts from rest. The ball has a mass of 2 kg and is placed 5 meters above the bottom of the ramp.
a) Calculate its potential energy
Halfway down the ramp: At some point during its descent, the ball will have a mix of potential energy and kinetic energy.
The ball is halfway down the ramp (h=2.5m).
b) Calculate its PE
c) Calculate its velocity
At the bottom : At the bottom of the ramp, all of the potential energy has been converted to kinetic energy.
d) Calculate its velocity
a) The total mechanical energy at the top is 98 J of potential energy.
b) The PE halfway down the ramp is 49J. The remaining 49 J of mechanical energy must now be in the form of KE.
c) Halfway down the ramp, the ball is moving at 7 m/s, and the total mechanical energy is still 98 J
d) The velocity of the ball is 9.8m/s.
Relationship between GPE and KE
At the highest point, the object has maximum potential energy and minimum (or zero) kinetic energy.
As the object moves downward, potential energy changes into kinetic energy, making the object speed up.
At the lowest point, the object has maximum kinetic energy and minimum potential energy
Total mechanical energy stays constant throughout the motion if there is no friction or air resistance
Effect of friction on a system
Causes a system to lose mechanical energy by converting some of it into heat. When friction is present, total mechanical energy decreases as energy is lost to heat
A car of mass 1,000 kg is moving at a speed of 20 m/s. What is its kinetic energy?
The car's kinetic energy is 200,000 J.
A cyclist and their bike have a combined mass of 80 kg. If their kinetic energy is 6,400 J, what is their velocity?
The cyclist's velocity is approximately 12.6 m/s.
A tennis ball is moving at a speed of 25 m/s with a kinetic energy of 312.5 J. What is the mass of the tennis ball?
The mass of the tennis ball is 1 kg
A 1,500 kg car is moving at 25 m/s and applies the brakes, coming to a stop after 50 meters. Assuming uniform deceleration, what is the car's deceleration, and how long does it take to stop?
v2= u2 + 2as
v= u+at
The car’s deceleration is 6.25 m/s² and it takes 4 seconds to stop.
A roller coaster car with a mass of 500 kg descends from a height of 40 meters. Assuming no friction, calculate its speed when it reaches the ground.
The roller coaster reaches a speed of 28 m/s at the bottom.
A 2 kg book is lifted to a height of 3 meters. What is the gravitational potential energy of the book?
The book’s gravitational potential energy is 58.8 J.
A hiker stands on top of a 200‐meter cliff. If the hiker's gravitational potential energy relative to the ground is 156,800 J, what is the hiker’s mass?
The hiker’s mass is 80 kg.
An apple weighing 0.2 kg is hanging from a tree. If the apple’s gravitational potential energy is 9.8 J, how high is the apple above the ground?
The apple is 5 meters above the ground.
A diver with a mass of 70 kg jumps off a 10‐meter‐high diving platform. Ignoring air resistance, what is the diver’s speed just before hitting the water?
The diver's speed is 14 m/s just before hitting the water
A 10 kg object is dropped from a height of 50 meters. How long will it take to hit the ground, and what is its velocity just before impact? Assume no air resistance.
v= u+at
It takes approximately 3.2 seconds for the object to hit the ground, and the velocity just before impact is 31.4 m/s.
Heat definition
The transfer of energy due to a temperature difference
Heat flows from high temperature to low temperature.
Heat vs temperature
Heat: Energy in transit. Temperature: A measure of thermal energy
Types of Heat Transfer
ConducƟon: The transfer of heat through direct contact between particles. Heat moves from the hotter object to the cooler object.
ConvecƟon: The transfer of heat in fluids (liquids or gases) through the movement of the fluid itself. Warm fluid rises, and cool fluid sinks, creating convecƟon currents.
RadiaƟon: The transfer of heat through electromagnetic waves (infrared radiation), without the requirement of a medium.
Why do metals feel colder than wood at the same temperature?
Even if both materials are at the same temperature (e.g., room temperature), metals feel colder because they are better conductors of heat compared to wood. Conductors (like metals) transfer heat away from your body more quickly than insulators (like wood), which makes them feel colder to the touch
How does heat transfer in a boiling pot of water?
In a boiling pot of water, convection is the primary form of heat transfer. The heat source (e.g., the stove) heats the water at the bottom of the pot. As the water heats up, it becomes less dense and rises to the top, while the cooler, denser water sinks to the bottom, creaƟng a convecƟon current. This cycle continues, causing the water to boil as heat is distributed throughout the pot.
How does radiation transfer heat from the sun to Earth?
Radiation is the transfer of heat through electromagnetic waves and doesn’t require a medium (such as air or water) to transfer energy. The sun emits infrared radiation, which travels through the vacuum of space and reaches Earth, warming the planet’s surface.
A metal rod is heated at one end. What type of heat transfer is occurring as the heat moves through the rod?
Conduction
Why does hot air rise and cold air sink in a room heated by a radiator?
Convection
Give an example of each type of heat transfer: conduction, convection, and radiation.
meat on bbq, boiling water, solar rays
If a 500 g block of copper is placed in a hot oven, what type of heat transfer occurs between the oven and the block of copper?
radiation, conduction and convection
Temperature definition
a measure of the average kinetic energy of the parƟcles in a substance.
Specific heat capacity definition
The rate at which a material changes temperature
Specific heat capacity (c)
J/kg°C (How many Joules of thermal energy will change the temperature of 1kg of the material by 1K.)
Materials with high specific heat (e.g., water) absorb a lot of heat energy without a large temperature change.
Specific heat capacity formula
Q=mcΔT
Q- heat energy (J)
m- mass
c-Specific heat capacity
ΔT- Temperature change (°C)
Calculate the energy required to heat 2 kg of water from 20°C to 80°C.
Specific heat capacity of water c = 4186 J/kg°C
Q=mcΔT
=2 x 4186 x (80-20)
=502 320J
How much energy is required to raise the temperature of 1 kg of water by 5°C?
Specific heat capacity of water is 4,186 J/kg°C
20,930 J
If 10,000 J of heat is applied to 0.5 kg of aluminium, and its temperature increases by 50°C, what is the specific heat capacity of aluminium?
The specific heat capacity of aluminium is 400 J/kg°C.
How much energy is required to raise the temperature of 2 kg of iron from 20°C to 120°C? Specific heat capacity of iron = 450 J/kg°C
90,000 J
A 500 g sample of water absorbs 10,000 J of energy. By how many degrees Celsius will the temperature of the water increase? Specific heat capacity of water is 4,186 J/kg°C
The temperature of the water will increase by approximately 4.78°C.
A 1 kg block of copper is heated to 100°C and then allowed to cool to room temperature (25°C). How much heat energy does it lose? Specific heat capacity of copper = 385 J/kg°C
The block of copper loses 28,875 J of energy
GPE, KE and Energy Conservation in Free Fall
in a closed system (without air resistance), the total mechanical energy remains constant. During free fall, gravitational potential energy (PE) at the start is completely converted into kinetic energy (KE) just before hitting the ground
Transformation of Energy in Free F
Gravitational potential energy (PE) at the start of the fall:
PE=mgh
Kinetic energy (KE) at the end of the fall (just before hitting the ground):
KE=1/2mv2
PEinitial = KEfinal
∴gh=1/2v2
A rock climber accidentally drops their climbing bag from a cliff at a height of 40 m. Calculate the bag’s velocity just before it hits the ground (ignore air resistance).
Given:
Gravity g =9.8 m/s2
Height h = 40 m
Step 1: Calculate velocity (mgh=1/2mv2)
The velocity of the bag just before hitting the ground is approximately 28.0 m/s.
Affect of mass on velocity during free fall
Regardless of their mass, the velocity at the point of impact is the same for all objects (ignoring air resistance), while the total amount of energy varies based on mass.
A stone is dropped from a cliff that is 20 meters high. What is the stone's velocity just before it hits the ground? (Assume no air resistance.)
v=19.8 m/s
An watermelon with a mass of 5 kg is dropped from a large crane at a height of 50 m. Calculate its PE at that height and determine its KE just before hitting the ground.
The watermelon’s kinetic energy just before hitting the ground is 2450J.
If a ball is thrown downward with an initial velocity of 5 m/s, from a height of 30 meters, what will be its velocity just before it hits the ground?
v2=u2 + as
v=24.8 m/s
A 10 kg object falls from rest and reaches a velocity of 29.4 m/s just before hitting the ground. Calculate the height from which the object was dropped
The object was dropped from a height of 44.1 m
An object is dropped from a height of 100 meters on Earth. What would its velocity be just before hitting the ground if the same object were dropped from the same height on the Moon, where gravity is 1.6 m/s
v=17.9 m/s
A rock climber drops a 12 kg backpack from a cliff 40m high.
a) Calculate the GPE of the backpack at the top.
b) Calculate the KE when it is 15m above the ground.
c) Calculate the velocity at 15m above the ground.
a)
GPE=mghtop
=12 x 9.8 x 40
=4704J
b)
GPE15m=mgh15m
=1764
Etotal=GPEtop
KE15m=Etotal-GPE15m
=4704-1764
=2940J t 15m above the ground
c) KE15m = 1/2 mv2
2940=1/2 x 12 x v2v15m=22.1m/s
A 0.5 kg water balloon is dropped from the roof of a building that is 15 meters high.
a) Calculate the GPE of the water balloon at the top.
b) Calculate the KE and velocity of the water balloon when it is 5 meters above the ground.
a) GPE = mgh = 0.5 × 9.8 × 15
= 73.5 J
b) GPE at 5m = mgh
= 0.5 × 9.8 × 5 = 24.5 J KE at 5m
= 73.5 - 24.5
= 49 J v²
= 2 × KE / m
= 2 × 49 / 0.5
= 196
v = 14 m/s
A bungee jumper with a mass of 60 kg jumps from a platform that is 80 meters above the ground. Assume the jumper is free-falling before the bungee cord stretches.
a) Calculate the GPE of the bungee jumper at the start.
b) Calculate the KE and velocity of the jumper when they have free fallen 50 meters.
a) GPE = mgh
= 60 × 9.8 × 80
= 47,040 J
b) GPE at 30m = mgh
= 60 × 9.8 × 30
= 17,640 J KE at 30m
= 47,040 - 17,640
= 29,400 J v²
= 2 × KE / m
= 2 × 29,400 / 60
= 980
v = 31.3 m/s
A ball is dropped from a table that is 1.5 m high. Calculate: - The potential energy of the ball at the top.
- The speed of the ball at a height of 0.5 m above the ground. Assume the ball’s mass is 0.3 kg and there is no air resistance.
PE = 4.41 J v ≈ 4.42 m/s
A 2 kg rock falls off a cliff that is 20 m high. Calculate: - The potential energy of the rock at the top. - The speed of the rock when it is halfway down. Assume no air resistance.
PE = 392 J v ≈ 14 m/s
An apple falls from a tree branch 5 m above the ground. If the apple’s speed is 8 m/s, and it’s KE is 6.4J when it is 2 m above the ground:
- Calculate the mass of the apple.
- Calculate its potential energy at 2 m height.
m ≈ 0.2 kg PE= 3.92 J
A coin with a mass of 0.05 kg falls from a table 1.5 m high. It hits the ground with a speed of 5.4 m/s.
- Calculate the gravitational acceleration, assuming no air resistance.
- Verify the potential energy of the coin at a height of 0.5 m above the ground.
g ≈ 9.8 m/s² PE = 0.245 J
An apple falls from a tree branch 5 m above the ground. If the apple’s speed is 8 m/s, and it’s KE is 6.4J when it is 2 m above the ground:- Calculate the mass of the apple. - Calculate its potential energy at 2 m height
m ≈ 0.2 kg PE= 3.92 J
A coin with a mass of 0.05 kg falls from a table 1.5 m high. It hits the ground with a speed of 5.4 m/s.
- Calculate the gravitational acceleration, assuming no air resistance.
- Verify the potential energy of the coin at a height of 0.5 m above the ground.
g ≈ 9.8 m/s² PE = 0.245 J
A 1.2 kg book falls off a shelf. If it gains 12 J of kinetic energy just before hitting the ground and the gravitational acceleration is 9.8 m/s²:
- Calculate the height of the shelf.
- Find the velocity of the book when it is 0.5 m from the ground.
h ≈ 1.02 m v ≈ 3.19 m/s
A 5 kg box is placed on a 30°, 8m long inclined plane. The box slides down the plane.
- Calculate the potential energy of the box at the top of the inclined plane. (Use g = 9.8 m/s²)
- Determine the velocity of the box when it reaches the bottom of the inclined plane.
1:
Since the inclined plane makes an angle of 30° with the horizontal, we can find the height (h) using sine:
sin(30°)=h / 8 m
h = 8 m × sin(30°)
= 8 m × 0.5
= 4 m
2: Potential Energy at Top
PE = mgh
Where:
- m = 5 kg
- g = 9.8 m/s²
- h = 4 m
PE = 5 × 9.8 × 4 = 196 J
3: Velocity at Bottom
kinetic energy (KE) is: KE = ½mv²
Since all the potential energy at the top is converted to kinetic energy at the bottom:
PE = KE 196 J = ½ × 5 × v²
solve for v:
v² = (2 × 196) / 5 v²
= 392 / 5
v² ≈ 78.4
v ≈ 8.85 m/s
Incline plane diagram
• Draw a clear, simple diagram of the inclined plane with the box at the top.
• Indicate the angle of the plane (30°) using a labeled arc.
• Represent forces acting on the box with arrows:
• Weight pointing straight down.
• Normal Force perpendicular to the plane.
• Component of Weight Parallel to the Plane pointing down the slope.
Label each force with a brief explanation. For example:
• Weight (mg): The gravitational force pulling the box down.
• Normal Force (N): The support force from the inclined plane.
• Parallel Force (mg sinθ): Causes the box to slide down. Add short notes explaining the relationships:
• “As the box slides, GPE converts to KE.”
A construction worker accidentally drops a 10 kg toolbox from a crane that is 25 meters above the ground.
a) Calculate the GPE of the toolbox at the top.
b) Calculate the KE and velocity of the toolbox when it is 10 meters above the ground.
a) GPE = mgh
= 10 × 9.8 × 25
= 2450 J
b) GPE at 10m = mgh
= 10 × 9.8 × 10
= 980 J KE at 10m
= 2450 - 980
= 1470 J v²
= 2 × KE / m
= 2 × 1470 / 10
= 294
v = 17.1 m/s
A skydiver with a mass of 75 kg jumps out of a plane at an altitude of 1000 meters.
a) Calculate the GPE of the skydiver at the start.
b) Calculate the kinetic KE and velocity of the skydiver when they have fallen 800 meters.
a) GPE = mgh
= 75 × 9.8 × 1000
= 735,000 J
b) GPE at 200m = mgh
= 75 × 9.8 × 200
= 147,000 J KE at 200m
= 735,000 - 147,000
= 588,000 J v²
= 2 × KE / m
= 2 × 588,000 / 75
= 15,680
v = 125.3 m/s
A rock climber drops a 12 kg backpack from a cliff 40m high
a) Calculate the GPE of the backpack at the top.
b) Calculate the KE when it is 15m above the ground.
c) Calculate the velocity at 15m above the ground.
Given: Mass of the backpack ( m) = 12 kg
Height of the cliff ( htop) = 40 m
Height from the ground to the point of interest ( h15m) = 15 m
Gravitational acceleration (g) = 9.8 m/s²
a) GPE=mghtop
=12 x 9.8 x 40
=4704J
b) GPE15m = mgh15m
=12 x 9.8 x 15
=1764J
Etotal=GPEtop
KE15m=Etotal=GPE15m
=4704-1764
=2940J at 15m above the ground
c) KE15m = 1/2mv2
2940 = 1/2 x 12 x v2
v15m = 22.1m/s
A 0.5 kg water balloon is dropped from the roof of a building that is 15 meters high.
a) Calculate the GPE of the water balloon at the top.
b) Calculate the KE and velocity of the water balloon when it is 5 meters above the ground.
a) GPE = mgh
= 0.5 × 9.8 × 15
= 73.5 J
b) GPE at 5m = mgh
= 0.5 × 9.8 × 5
= 24.5 J
KE at 5m = 73.5 - 24.5
49 J
v² = 2 × KE / m
= 2 × 49 / 0.5
= 196
v = 14 m/s
A bungee jumper with a mass of 60 kg jumps from a platform that is 80 meters above the ground. Assume the jumper is free-falling before the bungee cord stretches. a) Calculate the GPE of the bungee jumper at the start. b) Calculate the KE and velocity of the jumper when they have free fallen 50 meters.
a) GPE = mgh
= 60 × 9.8 × 80
= 47,040 J
b) GPE at 30m = mgh
= 60 × 9.8 × 30
= 17,640 J KE at 30m
= 47,040 - 17,640
= 29,400 J
v² = 2 × KE / m
= 2 × 29,400 / 60
= 980
v = 31.3 m/s
A ball is dropped from a table that is 1.5 m high. Calculate: - The potential energy of the ball at the top.
- The speed of the ball at a height of 0.5 m above the ground. Assume the ball’s mass is 0.3 kg and there is no air resistance.
PE = 4.41 J v ≈ 4.42 m/s
A 2 kg rock falls off a cliff that is 20 m high. Calculate: - The potential energy of the rock at the top. - The speed of the rock when it is halfway down. Assume no air resistance.
PE = 392 J v ≈ 14 m/s
An apple falls from a tree branch 5 m above the ground. If the apple’s speed is 8 m/s, and it’s KE is 6.4J when it is 2 m above the ground:
- Calculate the mass of the apple.
- Calculate its potential energy at 2 m height
m ≈ 0.2 kg PE= 3.92 J
A coin with a mass of 0.05 kg falls from a table 1.5 m high. It hits the ground with a speed of 5.4 m/s.
- Calculate the gravitational acceleration, assuming no air resistance.
- Verify the potential energy of the coin at a height of 0.5 m above the ground.
g ≈ 9.8 m/s² PE = 0.245 J
An apple falls from a tree branch 5 m above the ground. If the apple’s speed is 8 m/s, and it’s KE is 6.4J when it is 2 m above the ground:
- Calculate the mass of the apple.
- Calculate its potential energy at 2 m height.
m ≈ 0.2 kg PE= 3.92 J
A coin with a mass of 0.05 kg falls from a table 1.5 m high. It hits the ground with a speed of 5.4 m/s.
- Calculate the gravitational acceleration, assuming no air resistance.
- Verify the potential energy of the coin at a height of 0.5 m above the ground.
g ≈ 9.8 m/s² PE = 0.245 J
A 1.2 kg book falls off a shelf. If it gains 12 J of kinetic energy just before hitting the ground and the gravitational acceleration is 9.8 m/s²:
- Calculate the height of the shelf.
- Find the velocity of the book when it is 0.5 m from the ground.
h ≈ 1.02 m v ≈ 3.19 m/s
Equations for tension
F-vectors = T-mg
F-motion = ma
F-vectors = F-motion
Difference between speed and velocity
Speed is defined as the rate of change in distance with respect to time. Velocity is defined as the rate of change in displacement with respect to time.
Tension formula
T= mg+ma