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G481
MECHANICS MODULE 1:
WORK AND ENERGY
Answer Booklet
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RAB Plymstock School
Lesson 16 questions – Work Done
1.a) Travelling slowly, at about 5 miles an hour, the retarding forces acting on a cyclist are
about 5 N.
i)
What propulsive force must the cyclist provide to travel at constant speed?
Net force must be zero, so 5 N
Force = …5…………N (1)
ii)
How much energy does the cyclist have to provide to cover 5 m?
Work  force  distance
 5m 5N
 25 J
Energy = ……25……… J (2)
b)
At a higher speed, nearly 10 mph the retarding forces are much larger, nearly 8 N.
i)
Write down the propulsive force required at this speed.
Net force must be zero, so 8 N
Force = …8…………N (1)
ii)
Calculate the energy the cyclist must provide to cover 5 m at this constant speed.
Work  force  distance
 5m 8N
 40 J
Energy = ……40……… J (2)
c)
Climbing a hill at a steady speed of 5 mph the cyclist finds herself perspiring rather
more than on the flat. The mass of the cycle plus cyclist is 80 kg. The ergometer pedals fitted
report the energy supplied to the bicycle as 185 J to cover a 5 m stretch of road.
i)
How much energy is used to lift the cyclist uphill?
energy to lift  energy supplied  energy dissipated
 185 J  25 J
 160 J
Energy = ……160……………J (2)
ii)
How much height does she gain in travelling along the 5 m stretch of road?
Energy transferred  m  g  h
h 
160 J
80 N  10 N kg 1
 0 .2 m
Height = ……0.2……..m (3)
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iii)
How much energy would she have to provide to cover this same stretch of road at 10
mph?
energy supplied  energy to lift  energy dissipated
 160 J  40 J
 200 J
Energy = ……200……………. J (2)
iv).
What is the retarding force acting on her, when travelling uphill at 10 mph?
Energy transferre d  force  distance
force 
200 J
5m
 40 N
Force = …40…………….. N (2)
v)
Draw a vector diagram for the retarding forces (due to drag and gravity) acting on her at
10 mph.
drag
retarding force
due to gravity
(2)
d)i)
Calculate the retarding force due to the slope as a fraction of her weight.
ratio 
32 N
 4%
800 N
……4……% (2)
ii)
Comment on the advantages of using the slope of the hill rather than lifting bike and
rider vertically by the same distance.
………The retarding force as a percentage of weight was 4% whereas it would be 100% to go
vertically up. The power needed to travel the same distance vertically would be 25 times as
much because of this. …………………………………………………… (2)
Total [20]
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RAB Plymstock School
Lesson 27 questions – Energy calculations
Note: g = 9.8 N kg-1 = 9.8 m s-2
1
A jumper of mass 75 kg jumps a height of 1.9 m
(a)
Calculate the gravitational potential energy of the jumper at his highest point.
Gravitational potential energy = mgh = 75 x 9.8 x 1.9 = =1303.4 J = 1400 J to 2sf
gravitational potential energy = ……1400………… (J) (2)
(b)
Calculate the vertical velocity of the jumper just before impact with the crash mat.
v2 = u2 + 2as u = 0 so v  2gh. v = (2x 9.8 x 1.9)0.5 so v = 6.1 m s-1
(or from pe=ke)
vertical velocity just before impact = ………6.1…………… (ms-1) (3)
(c)
Show that the kinetic energy associated with his vertical velocity is equal to the
gravitational potential energy at the top of his jump. (ecf from b)
½ m v2 = 0.5 x 75 x 6.12 = 1395 ≈ 1400 J
QED
(2)
(d)
The jumper is stopped over a distance of 0.3 m when he lands on the crash mat.
Calculate the average force on the jumper as he comes to rest.
W=Fs
F=W/s
= 1400/0.3
Average force = ……4700………… (N) (2)
Total [10]
2a)
Explain the quantities
i)
Gravitational potential energy
……Energy transferred when moving an object of mass m a height h against a gravitational
field strength g.
or mgh where m – mass, g – gravitational field strength, h – height.
……………………………………………………………………………………………………
…………………………………………………………………………… (2)
ii)
kinetic energy
………energy due to movement
depending on mass and speed.
or …½ m v2 where m – mass and v
velocity/speed……………………………………………………………………… (2)
b)
Water leaves a reservoir and falls through a vertical height of 130m and causes a water
wheel to rotate. The rotating wheel is then used to produce 110 kJ per second of electrical
energy.
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RAB Plymstock School
i)
Calculate the velocity of the water as it reaches the wheel, assuming that all the
gravitational potential energy is converted to kinetic energy.
pe=ke
mgh=½ m v2
9.8 x 130 = 0.5 v2
v = (2x9.8x130)
velocity = ……50 (2sf)…………… ms-1 (3)
ii)
Calculate the mass of water flowing through the wheel per second, assuming the
production of electrical energy is 100% efficient.
electrical energy = 110 000 Js-1
110 = ½ m v2
m = (2x110000)/502
(or mgh)
mass of water per second = ……88………….. unit…kg…… (3)
iii)
State and explain two reasons why the mass of water flowing per second needs
to be greater than the value in (ii) in order to produce the amount of electrical energy stated.
………not all pe converted to ke through friction of water with ground or air resistance.
………not all ke transferred, the water retains some ke.
………friction in rotation of wheel so not all ke converted to electrical
………friction in generator not all ke converted to electrical
………………………………………………………………
……………………………………………………………………………………………………
……………………………………………………………………………………………………
……………………………………………………………………………… (2)
Total [12]
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RAB Plymstock School
Lesson 31 questions – Hooke’s Law
1
Fig 1.1 shows a spring that is fixed at one end and is hanging vertically.
ai)
Calculate the spring constant of the spring.
F/x=k
2.0N/0.25m
b)
spring constant = ……8……….. unit ……N/m…. (3)
The mass M is pulled down a further 150mm by a force F additional to its weight.
i)
Determine the force F.
extension of line
F = ……3.2………… N (1)
ii)
State any assumptions made.
…………elastic limit / ultimate tensile stress
not reached………………………………………………………………………………… (1)
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RAB Plymstock School
Lesson 32 questions - Strain Energy
1)a) State Hooke’s Law:
……Force is directly proportional to extension…………………………
………………………………………………………………………………………… (1)
b)
Fig 1.1 shows a graph of force F against extension e for a metal in the form of a wire.
The cross sectional area of the wire is 1.80x10-7 m2 and its length is 1.70m.
fig 1.1
i)
Calculate the Young modulus of the metal.
E=stress/strain
=(FL)/(AΔL)
=(29x1.70)/(1.80x10-7 x 1.6x10-3)
Young modulus = …1.71x1011………… Unit……Pa (Nm-2)……. (5)
ii)
Calculate the energy stored in the wire when it extended by 1.60mm.
W=1/2 FΔL
=1/2 x 29 x 1.6x10-3
energy stored = ……0.0232…………….. J (3)
Total [8]
2
a)
Hooke’s Law states that F=kx. What do F, k and x represent?
……F – Force applied to material
k – spring constant (stiffness)
x – extension of material…………………………………………… (1)
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RAB Plymstock School
b)
A spring is compressed by applying a force. Fig 2.1 shows the variation of force F with
compression x.
Fig 2.1
i)
Calculate the spring constant.
F=kx
F/x=k
grad = k
=125/50x10-3
spring constant = ……25000……….. unit ……Nm-1………. (2)
ii) Show that the work done in compressing the spring by 48mm is 2.9 J.
W=1/2 Fx
=1/2 x 120 x 48x10-3
=2.88≈2.9J
QED
(2)
c)
Fig 2.2 shows the spring in a toy gun. The spring is used to fire a dart of mass 15g
vertically.
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RAB Plymstock School
Fig 2.2
i)
The spring is compressed by 48mm in the gun. When the gun is fired the strain
energy in the spring is converted into kinetic energy of the dart. Calculate the speed with which
the dart initially leaves the spring when the gun is fired.
Assume all strain energy = kinetic energy
2.9J = ½ m v2
2.9 = 0.5 x (15x10-3)x v2
v=(2.9/(0.5 x (15x10-3)))
v=19.66
speed = …………19.7………….. ms-1 (3)
ii)
Give two reasons why the dart is unlikely to have 2.9J of gravitational potential
energy when it reaches its maximum height.
1……The dart will lose energy because of Air resistance …………………..
………………………………………………………………………………………………
………………………………………………………………………………………………
2………friction in the gun…………………………………………
………………………………………………………………………………………… (2)
Total [10]
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RAB Plymstock School
Lesson 33 questions - Young modulus
1
a)
A strip of rubber originally 75 mm long is stretched until it is 100 mm long.
What is the tensile strain?
strain 
extension
original length

25 mm
75 mm
This is sometimes expressed as a strain of 33%.
Strain = …… 0.33…….. (2)
b)
Why has the answer no units?
...........It is a ratio of two lengths.............................................................................
.....................................................................................................................................(1)
2.
The greatest tensile stress which steel of a particular sort can withstand without
breaking is about 109 N m-2. A wire of cross-sectional area 0.01 mm2 is made of this steel.
What is the greatest force that it can withstand?
stress 
force
area
so F = stress x A
= 109 N m–2
–8 m2
Force = ……10……. N (2)
3.
Find the minimum diameter of an alloy cable, tensile strength 75 MPa, needed to
support a load of 15 kN.
stress 
force
area
so
force
stress
15 kN
area 

75 MN m – 2
Area = 2 x 10–4 m2
A
d 2
4
so
d
d 2
4
d
4  2  10 4 m 2
3.14
diameter = …1.6 x 10–2…………unit…m…. (3)
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RAB Plymstock School
4.
Calculate the tensile stress in a suspension bridge supporting cable, of diameter of 50
mm, which pulls up on the roadway with a force of 4 kN.
A
d 2
4
3.14  (50  10 3 m)2
4

=1.96 x 10–3 m2
force
area
4 kN
stress 

1.96  10 -3 m2
Stress = ……2.0 …………. MPa
(3)
5.
Calculate the tensile stress in a nylon fishing line of diameter 0.36 mm which a fish is
pulling with a force of 20 N
A

d 2
4
3.14  (0.36  10 3 m)2
4
= 1.017 x 10-7 m2
stress 
force
area
20 N
1.017  10 – 7 m 2
Stress = 200 MPa

(3)
6.
A large crane has a steel lifting cable of diameter 36 mm. The steel used has a Young
modulus of 200 GPa. When the crane is used to lift 20 kN, the unstretched cable length is 25.0
m. Calculate the extension of the cable.
A
d 2
4
3.14  (3.6  10 2 m) 2
4
–3
=1.02 x 10 m2

E
F A
l l
so
l 

F l
A E
20  10 3 N  25 m
1.02  10 – 3 m 2  2  1011 N m – 2
Δl = 2.5 x 10–3 m
extension = …2.5 …………. unit … mm ……. (4)
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RAB Plymstock School
7
a) Define:
(i) stress
..........Force per unit cross section area.....................................................................
.....................................................................................................................................(1)
(ii) strain
.........extension per unit original length......................................................................
.....................................................................................................................................(1)
b)
(i) Distinguish between elastic and plastic behaviour when materials are stretched.
.......Elastic returns to original length/size.................................................................
........Plastic does not return to original length/size.....................................................
..........when load is released.......................................................................................
....................................................................................................................... (2)
(ii) Define elastic limit
.........Force (or extension) beyond which a material will not return to original
length......................................................................................................................
.....................................................................................................................................(1)
c)
(i) State the SI unit of the Youngs modulus.
unit =........Pascal (Nm-2)..............................(1)
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RAB Plymstock School
(ii) Describe, with the aid of a diagram, an experiment to determine the Youngs modulus of
steel in the form of a wire. Explain how to use your readings to obtain the Youngs modulus.
X
Y
.......... A load is attached to wire X to keep it taut. A load is attached to wire Y sufficient
to remove any kinks. (this is not counted as part of the final load)
The original length of Y is measured (1) from the support to the vernier scale with a
measuring tape or ruler. The diameter is measured with a micrometer, (1) (measured
at different points and averaged). From these, the mean radius is taken.
The initial vernier reading is noted with no load on Y. A load is then added and the
new vernier reading taken. This procedure is repeated adding loads in 10 Newton
steps up to a maximum of 60N. Loads are then removed and the vernier reading
recorded again. (1)
The extension for each load is calculated by subtracting the original vernier reading
from the mean reading.
A graph is then plotted of load F(N) against extension ΔL(m). (1)
The gradient = F/ΔL,
So E = gradient x L/πr2 (1)
The second wire makes sure that any temperature affects are eliminated. (1)
A long wire is used to increase the extension and reduce fractional uncertainty.
(reduce any errors made in measuring length and extension) (1)
(7)
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RAB Plymstock School
Lesson 34 questions – material properties
1)
Here is the graphical output from tensile testing of five materials: Perspex, polypropylene, aluminium,
brass and mild steel. Load is on the Y axis and extension is on the X axis. The extension is not to same scale as
they vary a lot.
Perspex
Polypropylene
4000
4000
3000
3000
2000
2000
1000
1000
0
0
extension
extension
Aluminium
Brass
2000
8000
1500
6000
1000
4000
500
2000
0
0
extension
extension
Mild steel
5000
4000
2000
1000
0
a)
State which material has the highest yield load and take a measurement of this value
……………… mild steel 4600 N (+/- 200N).……………………………………………… (2)
b)
Which sample has the lowest maximum ultimate stress?
………………Aluminium……………………………………………………………………….
……………………………………………………………………………………………… (1)
Total [3]
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RAB Plymstock School
2
Three materials P, Q and R have force-extension graphs as shown in Fig 2.1.
Fig 2.1
State and explain which of these three materials you would choose for the following
applications.
i)
making foundations of a building
……P…………………………………………………………………………………………….
……Has small extension for large compression forces and will return to initial
shape………………….
……………………………………………………………………………………………… (2)
ii)
forming into a wire
……Q………………………………………………………………………………………….
……under tension or compression it will return to initial shape but if enough force given it will
change shape and stay at this shape ……………………………………….
…………………………………………………………………………………………………….
……………………………………………………………………………………………… (2)
iii)
making an elastic rope (e.g. a bungee-jump rope)
……R…………………………………………………………………………………………….
……Will extend with small forces and seems to have polymeric shape, does not seem to have
reached elastic limit with forces shown meaning it will be able to go back to original
shape………………………………………………………………………………………… (2)
iv)
making a beam to support a roof
…… P…………………………………………………………………………………………….
……Has small extension for large compression forces and will return to initial shape … (2)
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RAB Plymstock School