Download 5 Forces Exam practice questions Pages 69

5 Forces
Exam practice questions
Pages 69-72 Exam practice questions
1 a) The weight is a downward gravitational force of the Earth on the man. By Newton’s third law
there must be an equal upward gravitational force of the man on the Earth – the answer is C.
[Total 1 Mark]
b) Only two forces act on the man: his weight pulling him down and the upward push of the
table on his feet. For the man to be in equilibrium these must be equal and opposite – the
answer is B.
[Total 1 Mark]
2 Examples of distant forces are: gravitational, electrostatic and magnetic
(electromagnetic) and nuclear.
[2]
Examples of contact forces are: friction, air resistance and normal reaction forces.
[2]
[Total 4 Marks]
3 Examples of gravitational forces are: Sun and planets, satellites, weight of objects on Earth
[1]
Examples of electromagnetic forces are: motors, electrons orbiting the nucleus, forces on static
electrical charges
[1]
Examples of nuclear forces are: strong forces binding protons and neutrons, weak forces involved
in beta decay
[1]
[Total 3 Marks]
4 a) The resultant force acting on the system = 7000 N − (1000 + 1000) N
= 5000 N.
Using Newton’s second law: a 
F 5000 N

 1.25 m s 2 – the answer is A
m 4000 kg
[Total 1 Mark]
b) Resultant force on the trailer = (T − 1000) N = 2500 kg × 1.00 m s−2
T = 2500 N + 1000 N = 3500 N – the answer is B
[Total 1 Mark]
5 a) A body is in equilibrium if the vector sum of all the forces acting on it is zero.
[1]
b) The student needs:
• A diagram showing weights and hangers and pulley wheels (see figure 5.5)
[1]
• Pencil lines drawn to show the direction of the two outer forces
[1]
• To choose a scale and mark off the appropriate lengths of the force lines
[1]
• To construct a parallelogram
[1]
• To measure the length of the diagonal to determine the magnitude
of the resultant of the two forces
[1]
• The resultant should be equal and opposite to the third weight for the system to be in
equilibrium
[1]
[Total 7 Marks]
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5 Forces
Exam practice questions
6 a) Free body force diagram should include all of the 4 forces stated.
b) Weight
[1]
c) i)
[1]
F = (4.0 kN) cos 15
= 3.86 kN ≈ 3.9 kN
ii)
R + (4.0 kN) sin 15 = 800 kg × 9.8 m s−2
[1]
⇒ R = 6.81 kN ≈ 7 kN
[1]
[Total 8 Marks]
7 a) i) [1] and ii) [1]
b)
[1]
c) Diagram [1]; R = mg tan 30 ° [1] = 0.28 N [1]; or by scale drawing (± 0.02 N) [3]
[Total 6 Marks]
8 a) Newton’s third law states that if body A exerts a force on body B, then body B will exert an
equal and opposite force on body A.
b) i)
[1]
Only one force on each body i.e a force from the centre of the Moon towards the centre
of the Earth and a force from the centre of the Earth towards the centre of the Moon. [2]
ii)
The forces are both gravitational, of the same magnitude, co-linear and act in opposite
directions. (any three)
[3]
[Total 6 Marks]
9 a) The nozzle exerts a force on the water to shoot it out of the rear of the jet-ski [1], so the
water will exert an equal and opposite force on the jet-ski [1], pushing it forwards.
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5 Forces
Exam practice questions
b) F  v
F v
m
t
m
t
= 5.0 m s−1 × 150 kg s−1 = 750 N
[1]
= 6.0 m s−1 × 160 kg s−1 = 960 N
[1]
c) The resistive forces in part (b) must equal the driving force for the velocity to be constant.
Assuming that these remain the same during the acceleration [1], the resultant force on the
jet-ski will be:
FR = 960 N – 750 N = 210 N
Hence acceleration, a =
F
[1]
=
m
210 N
380 kg
= 0.56 m s−2
[1]
Using v = u + at, final velocity = 15 m s−1 + 0.56 m s−2 × 5.0 s = 18 m s−1
[1]
[Total 8 Marks]
10 a) N + F – W = 0; or N + F = W
[1]
b) 8.0 kg × 9.8 m s−2 × 0.10 m = F × 0.50 m [1]
F = 16 N
[1]
c) Moving the centre of gravity to a lower position would reduce the perpendicular distance of
the line of action of the weight from the pivot [1] and so reduce the clockwise moment. The
value of the anticlockwise moment will also reduce [1] and so the size of force F needs to be
less to maintain equilibrium.
[1]
[Total 6 Marks]
11 a) Resultant force = 4000 N – (2250 + 500) N = 1250 N [1] = 3750 kg × a
a=
1250 N
3750 kg
[1] = 0.33 m s−2
[1]
b) For the plough alone :T – 2250 N = 1250 kg × 0.33 m s−2
[1]
T = 2670 N
[1]
c) Coupling force = 500 N [1]; assuming that the resistive force has remained constant
[1]
[Total 7 Marks]
12 a) The force on truck B is given by the rate of change of momentum of truck B.
FB 
 ( mv ) B
t

45000 kg m s
1
 15000 kg m s
1
[2]
(1.0 s  0.64 s)
 FB = 83 000 kg m s−2 or 83 000 N
[1]
(b) The force on truck A is given by the rate of change of momentum of truck A.
FA 
 ( mv ) A
t

30000 kg m s
1
 60000 kg m s
1.0 s  0.64 s
 FA = −83 000 kg m s−2 or −83 000 N
1
[2]
[1]
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5 Forces
Exam practice questions
(c) The two forces are equal in size but opposite in direction, as we would expect from Newton’s
third law.
[2]
13 The argument is not valid because Newton’s third law cannot be applied to a single body.
[1]
There are two pairs of forces involving the book:
1 the downward gravitational pull of the Earth on the book and the upward gravitational pull of
the book on the Earth
[1]
2 the elastic contact force of the book pushing down on the table and the elastic contact force
of the table pushing upwards on the book
[1]
The book is in equilibrium because the forces acting on the book alone (the downward
gravitational force of the Earth and the upward elastic force of the table) are equal and
opposite
[1].
The resultant force acting on the book is therefore zero and so it is in equilibrium.
[1]
[Total 5 Marks]
14 a) The Earth pulls the teacher plus suitcase down with a gravitational force of 800 N [1]; the
teacher/case pull the Earth upward with a gravitational force of 800 N.
[1]
The suitcase pulls down on the spring with a force of 98 N [1]; the spring pulls the suitcase up
with an elastic force of 98 N
[1]
b)
To accelerate upwards the teacher and case need a resultant upward force FR = upward push
of scales – weight of teacher plus case = ma
Scale reading is equal to upward push = ma + mg =m (a + g) = 82 kg × (9.8 + 2.0) m s−2 = 970 N[1]
Spring balance reading = 10 kg × (9.8 + 2.0) m s−2 = 118 N
[1]
At constant velocity the resultant force is zero
Scale reading = 800 N [1]
Spring balance reading = 98 N
[1]
Using v2 = u2 + 2as for the initial stage
v2= 0 + 2 × 2 m s−2 × 4 m → v = 4.0 m s−1
[1]
Using v2 = u2 + 2as for the final stage
0 = (4.0 m s−1)2 – 2 × a × 8 m → a = −1 m s−2
[1]
When slowing down there is a resultant downward force = weight – scale reading
Scale reading = mg –ma =82 kg × (9.8 − 1.0) m s−2 =720 N
−2
Spring balance reading = 10 kg × (9.8 − 1.0) m s = 88 N
[1]
[1]
[Total 12 Marks]
15 a) Using v2 = u2 + 2as
b)
c)
v2 = 0 + 2 × 9.8 m s−2 × 0.20 m
→ v = 2.0 m s−1
[1]
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5 Forces
Exam practice questions
d) F 
mv
t
=
0.020 kg  2.0 m s
1
0.50 s
-0
(1) = 0.080 N
[1]
e) When the sphere enters the water it experiences an upward force from the water [1]. By
Newton’s third law the sphere will exert an equal downward push on the water [1] that will
result in an increase in the reading on the balance. The average force of the water on the
sphere is 0.080 N, which would show an increase of about 80 g [1] on the balance, but,
because this is an average value, the balance reading will momentarily show a higher value
during the first 0.50 s [1].
At the instant that the sphere comes to rest in the water it will be experiencing an ‘upthrust’
from the water [1]. This will produce an equal downward push on the water [1]. The upthrust
must be greater than the weight of the sphere because it causes it to move up towards the
surface. The balance reading at this point will be greater than 820 g [1].
As the sphere moves upwards there will be downward resistive forces [1] that will reduce the
resultant upward force by the water on the ball and so the balance reading will fall [1].
When the sphere is floating, the upthrust equals the weight of the sphere [1]. The sphere
therefore pushes down on the water with a force equal to its weight and so the balance will
give a steady reading of 820 g [1].
[Any 6 marks]
The beaker could be placed on a pressure/force sensor interfaced with a computer. Suitable
software could be used to take readings of pressure or force at very short time intervals and a
display of force against time produced on the screen.
[1]
[Total 10 Marks]
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