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GRADE 9 PRACTICE

Physics Problems: Curvilinear movement and balance, Grade 9

Centripetal force, satellites and stability of bodies.

18 problems across 4 sub-topics
Practise in the interactive mode

Here the conditions and analysis of the solution can be read without an account. In the simulator, you need to collect the same analyzes yourself: “Given,” formulas, order of steps and answer.

Problem statements

The answer options are there to check against, and the walkthrough opens under each problem.

18 problems
Position in this set: 1 from 18

Problem #16

Circular dynamics and satellites

What concept does the description correspond to: “This is the name for the resultant of forces directed towards the center and creating centripetal acceleration.”?

Answer options

  1. \text{Centripetal force}
  2. \text{Artificial satellite}
  3. \text{First cosmic speed}
  4. \text{Point particle}
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Given
\text{Description}
\text{The net force directed towards the centre and producing centripetal acceleration.}
Find

\text{name concepts}

  1. \text{Let's compare approval with definition}

    This is not a new kind of force: gravitation, tension, friction or their sum can fulfil this role.

  2. \text{Let's formulate conclusion}\text{Concept} = \text{Centripetal force}

Answer\text{Concept} = \text{Centripetal force}

Position in this set: 2 from 18

Problem #17

Circular dynamics and satellites

What concept does the description correspond to: “A body moves in orbit when gravity continuously imparts acceleration to it towards the center of the planet.”?

Answer options

  1. \text{Artificial satellite}
  2. \text{Centripetal force}
  3. \text{First cosmic speed}
  4. \text{Point particle}
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Given
\text{Description}
\text{A body moves in orbit when gravitation continuously gives it acceleration towards the planet's centre.}
Find

\text{name concepts}

  1. \text{Let's compare approval with definition}

    The satellite remains in free fall, continually missing the surface because of its horizontal velocity.

  2. \text{Let's formulate conclusion}\text{Concept} = \text{Artificial satellite}

Answer\text{Concept} = \text{Artificial satellite}

Position in this set: 3 from 18

Problem #18

Gravity

Estimate the first escape velocity of the Earth at g = 9.8 m/s² and a radius of 6.4·10⁶ m.

Answer options

  1. 7{,}9\,\text{км/с}
  2. 0{,}79\,\text{км/с}
  3. 9{,}8\,\text{км/с}
  4. 79\,\text{км/с}
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Given
g
9{,}8\,\text{м/с}^{2}
R
6{,}4\cdot10^6\,\text{м}
Find

v_1

  1. v_1=\sqrt{gR}v_1=\sqrt{9{,}8\cdot6{,}4\cdot10^6}\,\text{м/с}v_1 = 7{,}92\cdot10^3\,\text{м/с}
  2. v_1\approx7{,}9\,\text{км/с}

Answer7{,}9\,\text{км/с}

Position in this set: 4 from 18

Problem #19

Equilibrium and centre of gravity

What concept does the description correspond to: “The point of application of the resultant forces of gravity acting on parts of the body in a uniform field.”?

Answer options

  1. \text{Centre of gravity}
  2. \text{Conditions for equilibrium}
  3. \text{Stability of equilibrium}
  4. \text{Point particle}
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Given
\text{Description}
\text{The point where the resultant gravitational force on a body's parts acts in a uniform field.}
Find

\text{name concepts}

  1. \text{Let's compare approval with definition}

    Its position can be found by suspension or symmetry; for a hollow body it may lie outside the material.

  2. \text{Let's formulate conclusion}\text{Concept} = \text{Centre of gravity}

Answer\text{Concept} = \text{Centre of gravity}

Position in this set: 5 from 18

Problem #20

Equilibrium and centre of gravity

What concept does the description correspond to: “For a body to be in equilibrium, both the resultant forces and the total moment of forces relative to the axis are equal to zero.”?

Answer options

  1. \text{Conditions for equilibrium}
  2. \text{Centre of gravity}
  3. \text{Stability of equilibrium}
  4. \text{Point particle}
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Given
\text{Description}
\text{For a body in equilibrium, both the net force and the total moment about the axis are zero.}
Find

\text{name concepts}

  1. \text{Let's compare approval with definition}

    The first condition rules out translational acceleration; the second rules out angular acceleration.

  2. \text{Let's formulate conclusion}\text{Concept} = \text{Conditions for equilibrium}

Answer\text{Concept} = \text{Conditions for equilibrium}

Position in this set: 6 from 18

Problem #21

Statics

A force of 80 N acts on a 0.3 m arm. What force on the opposite 0.6 m arm can balance the lever?

Answer options

  1. 40\,\text{Н}
  2. 20\,\text{Н}
  3. 80\,\text{Н}
  4. 160\,\text{Н}
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Given
F_1
80\,\text{Н}
l_1
0{,}3\,\text{м}
l_2
0{,}6\,\text{м}
Find

F_2

  1. F_1l_1=F_2l_2
  2. F_2=\frac{F_1l_1}{l_2}F_2=\frac{80\cdot0{,}3}{0{,}6}\,\text{Н}F_2 = 40\,\text{Н}

Answer40\,\text{Н}

Position in this set: 7 from 18

Problem #94

Circular dynamics and satellites

An object moves uniformly in a circle of radius 5 m at speed 10 m/s. Find the acceleration.

Answer options

  1. 20\,\text{м/с}^{2}
  2. 40\,\text{м/с}^{2}
  3. 10\,\text{м/с}^{2}
  4. 60\,\text{м/с}^{2}
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Given
r
5\,\text{м}
v
10\,\text{м/с}
Find

a

  1. a=\frac{v^2}{r}a=\frac{{10}^{2}}{5}a=20\,\text{м/с}^{2}

    Even with a constant velocity module, its direction changes.

Answer20\,\text{м/с}^{2}

Position in this set: 8 from 18

Problem #95

Circular dynamics and satellites

An object moves uniformly in a circle of radius 8 m at speed 12 m/s. Find the acceleration.

Answer options

  1. 36\,\text{м/с}^{2}
  2. 18\,\text{м/с}^{2}
  3. 9\,\text{м/с}^{2}
  4. 54\,\text{м/с}^{2}
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Given
r
8\,\text{м}
v
12\,\text{м/с}
Find

a

  1. a=\frac{v^2}{r}a=\frac{{12}^{2}}{8}a=18\,\text{м/с}^{2}

    Even with a constant velocity module, its direction changes.

Answer18\,\text{м/с}^{2}

Position in this set: 9 from 18

Problem #96

Circular dynamics and satellites

Body mass 2 kg moves uniformly in a circle of radius 4 m at speed 6 m/s. Find the modulus of the resultant.

Answer options

  1. 36\,\text{Н}
  2. 18\,\text{Н}
  3. 9\,\text{Н}
  4. 54\,\text{Н}
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Given
m
2\,\text{кг}
r
4\,\text{м}
v
6\,\text{м/с}
Find

F

  1. a=\frac{v^2}{r}a=\frac{{6}^{2}}{4}a=9\,\text{м/с}^{2}

    The resultant is directed towards the center.

  2. F=maF=2 \cdot 9F=18\,\text{Н}

    Centripetal force is not a separate type of force, but a resultant one.

Answer18\,\text{Н}

Position in this set: 10 from 18

Problem #97

Circular dynamics and satellites

Body mass 3 kg moves uniformly in a circle of radius 6 m at speed 8 m/s. Find the modulus of the resultant.

Answer options

  1. 64\,\text{Н}
  2. 16\,\text{Н}
  3. 32\,\text{Н}
  4. 96\,\text{Н}
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Given
m
3\,\text{кг}
r
6\,\text{м}
v
8\,\text{м/с}
Find

F

  1. a=\frac{v^2}{r}a=\frac{{8}^{2}}{6}a=10{,}6667\,\text{м/с}^{2}

    The resultant is directed towards the center.

  2. F=maF=3 \cdot 10{,}6667F=32\,\text{Н}

    Centripetal force is not a separate type of force, but a resultant one.

Answer32\,\text{Н}

Position in this set: 11 from 18

Problem #98

Circular dynamics and satellites

Radius of the planet 6000 km, the satellite moves at an altitude 400 km. In this orbit, the acceleration due to gravity 9 m/s². Find the orbital speed in m/s for a circular orbit.

Answer options

  1. 15178{,}9\,\text{м/с}
  2. 3794{,}73\,\text{м/с}
  3. 7589{,}47\,\text{м/с}
  4. 22768{,}4\,\text{м/с}
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Given
R
6000\,\text{км}
h
400\,\text{км}
g
9\,\text{м/с}^{2}
Find

v

  1. r=(R+h)1000r=(6000 + 400) \cdot 1000r=6400000\,\text{м}

    The orbital radius is measured from the center of the planet.

  2. v^2=grv^2=9 \cdot 6400000v^2=57600000\,\text{м}^{2}\text{/с}^{2}

    Gravity provides centripetal acceleration.

  3. v=\sqrt{gr}v=\sqrt{57600000}v=7589{,}47\,\text{м/с}

    We take the positive root.

Answer7589{,}47\,\text{м/с}

Position in this set: 12 from 18

Problem #99

Circular dynamics and satellites

Radius of the planet 6000 km, the satellite moves at an altitude 1000 km. In this orbit, the acceleration due to gravity 7 m/s². Find the orbital speed in m/s for a circular orbit.

Answer options

  1. 14000\,\text{м/с}
  2. 3500\,\text{м/с}
  3. 21000\,\text{м/с}
  4. 7000\,\text{м/с}
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Given
R
6000\,\text{км}
h
1000\,\text{км}
g
7\,\text{м/с}^{2}
Find

v

  1. r=(R+h)1000r=(6000 + 1000) \cdot 1000r=7000000\,\text{м}

    The orbital radius is measured from the center of the planet.

  2. v^2=grv^2=7 \cdot 7000000v^2=49000000\,\text{м}^{2}\text{/с}^{2}

    Gravity provides centripetal acceleration.

  3. v=\sqrt{gr}v=\sqrt{49000000}v=7000\,\text{м/с}

    We take the positive root.

Answer7000\,\text{м/с}

Position in this set: 13 from 18

Problem #100

Equilibrium and centre of gravity

Force 20 H is applied perpendicular to the lever at a distance 0{,}3 m from the axis. Find the moment of force.

Answer options

  1. 12\,\text{Н·м}
  2. 6\,\text{Н·м}
  3. 3\,\text{Н·м}
  4. 18\,\text{Н·м}
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Given
F
20\,\text{Н}
l
0{,}3\,\text{м}
Find

M

  1. M=FlM=20 \cdot 0{,}3M=6\,\text{Н·м}

    The shoulder here is equal to the distance from the axis to the point of application.

Answer6\,\text{Н·м}

Position in this set: 14 from 18

Problem #101

Equilibrium and centre of gravity

Force 30 H is applied perpendicular to the lever at a distance 0{,}4 m from the axis. Find the moment of force.

Answer options

  1. 24\,\text{Н·м}
  2. 6\,\text{Н·м}
  3. 12\,\text{Н·м}
  4. 36\,\text{Н·м}
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Given
F
30\,\text{Н}
l
0{,}4\,\text{м}
Find

M

  1. M=FlM=30 \cdot 0{,}4M=12\,\text{Н·м}

    The shoulder here is equal to the distance from the axis to the point of application.

Answer12\,\text{Н·м}

Position in this set: 15 from 18

Problem #102

Equilibrium and centre of gravity

On the shoulder 10 cm force acts 60 N. What counter-rotating force should be applied to the shoulder? 30 cm for balance? Neglect the mass of the lever.

Answer options

  1. 40\,\text{Н}
  2. 10\,\text{Н}
  3. 20\,\text{Н}
  4. 60\,\text{Н}
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Given
l_1
10\,\text{см}
F_1
60\,\text{Н}
l_2
30\,\text{см}
Find

F_2

  1. M=F_1 l_1M=60 \cdot 10M=600\,\text{Н·см}

    We calculate the moment of the first force in agreed units.

  2. F_2=\frac{M}{l_2}F_2=\frac{600}{30}F_2=20\,\text{Н}

    For equilibrium, opposite moments are equal.

Answer20\,\text{Н}

Position in this set: 16 from 18

Problem #103

Equilibrium and centre of gravity

On the shoulder 15 cm force acts 80 N. What counter-rotating force should be applied to the shoulder? 40 cm for balance? Neglect the mass of the lever.

Answer options

  1. 60\,\text{Н}
  2. 15\,\text{Н}
  3. 90\,\text{Н}
  4. 30\,\text{Н}
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Given
l_1
15\,\text{см}
F_1
80\,\text{Н}
l_2
40\,\text{см}
Find

F_2

  1. M=F_1 l_1M=80 \cdot 15M=1200\,\text{Н·см}

    We calculate the moment of the first force in agreed units.

  2. F_2=\frac{M}{l_2}F_2=\frac{1200}{40}F_2=30\,\text{Н}

    For equilibrium, opposite moments are equal.

Answer30\,\text{Н}

Position in this set: 17 from 18

Problem #104

Equilibrium and centre of gravity

To the left of the axis of the weightless lever there are weights hanging 2 kg at a distance 0{,}3 m and 1 kg at a distance 0{,}6 m. At what distance to the right should the load be hung 4 kg? g = 10 N/kg.

Answer options

  1. 0{,}6\,\text{м}
  2. 0{,}15\,\text{м}
  3. 0{,}9\,\text{м}
  4. 0{,}3\,\text{м}
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Given
m_1
2\,\text{кг}
l_1
0{,}3\,\text{м}
m_2
1\,\text{кг}
l_2
0{,}6\,\text{м}
m_3
4\,\text{кг}
g
10\,\text{Н/кг}
Find

l

  1. M_1=m_1 gl_1M_1=2 \cdot 10 \cdot 0{,}3M_1=6\,\text{Н·м}

    The first load creates a moment about the axis.

  2. M_2=m_2 gl_2M_2=1 \cdot 10 \cdot 0{,}6M_2=6\,\text{Н·м}

    Both left moments are directed equally.

  3. l=\frac{M_1+M_2}{m_3g}l=\frac{(6 + 6)}{(4 \cdot 10)}l=0{,}3\,\text{м}

    The right moment balances the sum of the left ones.

Answer0{,}3\,\text{м}

Position in this set: 18 from 18

Problem #105

Equilibrium and centre of gravity

To the left of the axis of the weightless lever there are weights hanging 3 kg at a distance 0{,}2 m and 2 kg at a distance 0{,}3 m. At what distance to the right should the load be hung 6 kg? g = 10 N/kg.

Answer options

  1. 0{,}2\,\text{м}
  2. 0{,}4\,\text{м}
  3. 0{,}1\,\text{м}
  4. 0{,}6\,\text{м}
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Given
m_1
3\,\text{кг}
l_1
0{,}2\,\text{м}
m_2
2\,\text{кг}
l_2
0{,}3\,\text{м}
m_3
6\,\text{кг}
g
10\,\text{Н/кг}
Find

l

  1. M_1=m_1 gl_1M_1=3 \cdot 10 \cdot 0{,}2M_1=6\,\text{Н·м}

    The first load creates a moment about the axis.

  2. M_2=m_2 gl_2M_2=2 \cdot 10 \cdot 0{,}3M_2=6\,\text{Н·м}

    Both left moments are directed equally.

  3. l=\frac{M_1+M_2}{m_3g}l=\frac{(6 + 6)}{(6 \cdot 10)}l=0{,}2\,\text{м}

    The right moment balances the sum of the left ones.

Answer0{,}2\,\text{м}

The theory behind these problems

Work through the topic if a problem will not give in.

Circular dynamics and satellitesCentripetal force and first cosmic speed.Equilibrium and centre of gravityMoments of forces, centre of gravity and stability of equilibrium.

Formulas by topic

Look up symbols and units of measurement in the reference.

Torque (moment of a force)M · Н·мForceF · НVelocityv · м/с