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GRADE 7 PRACTICE

Physics Problems: Work, power and energy, Grade 7

Work, simple mechanisms, efficiency and energy conversion.

27 problems across 6 sub-topics
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Problem statements

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27 problems
Position in this set: 1 from 27

Problem #49

Mechanical work and power

What concept does the description correspond to: “A force does mechanical work when an object is displaced and the force has a component along that displacement.»?

Answer options

  1. \text{Mechanical work}
  2. \text{Power}
  3. \text{Conditions for doing work}
  4. \text{Physical phenomenon}
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Given
\text{Description}
\text{A force does mechanical work when an object is displaced and the force has a component along that displacement.}
Find

\text{name concepts}

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

    For a constant force in the direction of motion, work equals force times distance; without displacement, mechanical work is zero.

  2. \text{Let's formulate conclusion}\text{Concept} = \text{Mechanical work}

Answer\text{Concept} = \text{Mechanical work}

Position in this set: 2 from 27

Problem #50

Mechanical work and power

What concept does the description correspond to: “The physical quantity equal to work done per unit time, describing the rate of doing work.»?

Answer options

  1. \text{Power}
  2. \text{Mechanical work}
  3. \text{Conditions for doing work}
  4. \text{Physical phenomenon}
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Given
\text{Description}
\text{The physical quantity equal to work done per unit time, describing the rate of doing work.}
Find

\text{name concepts}

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

    Two machines can do the same work, but the more powerful one does it in less time.

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

Answer\text{Concept} = \text{Power}

Position in this set: 3 from 27

Problem #51

Work and power

A winch pulls a load with a constant force of 300 N over 4 m along the force in 20 s. Find the work and average power.

Answer options

  1. 1200\,\text{Дж};\ 60\,\text{Вт}
  2. 75\,\text{Дж};\ 15\,\text{Вт}
  3. 1200\,\text{Дж};\ 6000\,\text{Вт}
  4. 6000\,\text{Дж};\ 300\,\text{Вт}
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Given
F
300\,\text{Н}
s
4\,\text{м}
t
20\,\text{с}
Find

AN

  1. A=FsA=300\cdot4\,\text{Дж}A = 1200\,\text{Дж}
  2. N=\frac{A}{t}N=\frac{1200}{20}\,\text{Вт}N = 60\,\text{Вт}

Answer1200\,\text{Дж}60\,\text{Вт}

Position in this set: 4 from 27

Problem #52

Levers and pulleys

What concept does the description correspond to: “A rigid object that can rotate about a support and transmit the effects of forces with different moment arms.»?

Answer options

  1. \text{Lever}
  2. \text{Pulleys and the golden rule}
  3. \text{Torque (moment of a force)}
  4. \text{Physical phenomenon}
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Given
\text{Description}
\text{A rigid object that can rotate about a support and transmit the effects of forces with different moment arms.}
Find

\text{name concepts}

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

    A longer moment arm lets a smaller force balance a larger force on a shorter arm.

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

Answer\text{Concept} = \text{Lever}

Position in this set: 5 from 27

Problem #53

Levers and pulleys

What concept does the description correspond to: “Simple machines gain force at the cost of extra distance and ideally do not reduce the work required.»?

Answer options

  1. \text{Pulleys and the golden rule}
  2. \text{Lever}
  3. \text{Torque (moment of a force)}
  4. \text{Physical phenomenon}
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Given
\text{Description}
\text{Simple machines gain force at the cost of extra distance and ideally do not reduce the work required.}
Find

\text{name concepts}

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

    A fixed pulley changes force direction; an ideal movable pulley halves the required force.

  2. \text{Let's formulate conclusion}\text{Concept} = \text{Pulleys and the golden rule}

Answer\text{Concept} = \text{Pulleys and the golden rule}

Position in this set: 6 from 27

Problem #54

Simple mechanisms

The 0.4 m arm of the lever is subject to a force of 50 N. What force must be applied to the opposite 0.1 m arm for balance?

Answer options

  1. 200\,\text{Н}
  2. 12{,}5\,\text{Н}
  3. 50\,\text{Н}
  4. 500\,\text{Н}
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Given
F_1
50\,\text{Н}
l_1
0{,}4\,\text{м}
l_2
0{,}1\,\text{м}
Find

F_2

  1. F_1l_1=F_2l_2
  2. F_2=\frac{F_1l_1}{l_2}F_2=\frac{50\cdot0{,}4}{0{,}1}\,\text{Н}F_2 = 200\,\text{Н}

Answer200\,\text{Н}

Position in this set: 7 from 27

Problem #55

Efficiency and mechanical energy

What concept does the description correspond to: “The ratio of useful work to total input work, usually expressed as a percentage.»?

Answer options

  1. \text{Efficiency}
  2. \text{Potential energy}
  3. \text{Kinetic energy}
  4. \text{Physical phenomenon}
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Given
\text{Description}
\text{The ratio of useful work to total input work, usually expressed as a percentage.}
Find

\text{name concepts}

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

    Friction and deformation make a real machine's efficiency less than one hundred percent, but energy is dissipated rather than destroyed.

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

Answer\text{Concept} = \text{Efficiency}

Position in this set: 8 from 27

Problem #56

Efficiency and mechanical energy

What concept does the description correspond to: “Interaction energy that depends on the relative positions of objects or the deformation of an elastic object.»?

Answer options

  1. \text{Potential energy}
  2. \text{Efficiency}
  3. \text{Kinetic energy}
  4. \text{Physical phenomenon}
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Given
\text{Description}
\text{Interaction energy that depends on the relative positions of objects or the deformation of an elastic object.}
Find

\text{name concepts}

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

    A raised load can do work as it descends; a compressed spring can do work as it returns to its original shape.

  2. \text{Let's formulate conclusion}\text{Concept} = \text{Potential energy}

Answer\text{Concept} = \text{Potential energy}

Position in this set: 9 from 27

Problem #57

Efficiency and energy

The lift spent 1000 J of work, of which 800 J went to lifting the load. Find the efficiency.

Answer options

  1. 80\,\%
  2. 20\,\%
  3. 125\,\%
  4. 800\,\%
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Given
A_{\text{пол}}
800\,\text{Дж}
A_{\text{затр}}
1000\,\text{Дж}
Find

\eta

  1. \eta=\frac{A_{\text{пол}}}{A_{\text{затр}}}\eta=\frac{800}{1000}\eta = 0{,}8
  2. \eta_{\%}=0{,}8\cdot100\,\%\eta_{\%} = 80\,\%

Answer80\,\%

Position in this set: 10 from 27

Problem #154

Mechanical work and power

Force 40 H is directed along the movement 5 m. Find the work done by the force.

Answer options

  1. 200\,\text{Дж}
  2. 400\,\text{Дж}
  3. 100\,\text{Дж}
  4. 600\,\text{Дж}
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Given
F
40\,\text{Н}
S
5\,\text{м}
Find

A

  1. A=FSA=40 \cdot 5A=200\,\text{Дж}

    Force and displacement are co-directed.

Answer200\,\text{Дж}

Position in this set: 11 from 27

Problem #155

Mechanical work and power

Force 60 H is directed along the movement 8 m. Find the work done by the force.

Answer options

  1. 960\,\text{Дж}
  2. 480\,\text{Дж}
  3. 240\,\text{Дж}
  4. 1440\,\text{Дж}
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Given
F
60\,\text{Н}
S
8\,\text{м}
Find

A

  1. A=FSA=60 \cdot 8A=480\,\text{Дж}

    Force and displacement are co-directed.

Answer480\,\text{Дж}

Position in this set: 12 from 27

Problem #156

Mechanical work and power

Load weight 20 kg evenly raised by 6 m for 12 s. Find the useful power at g = 10 N/kg.

Answer options

  1. 200\,\text{Вт}
  2. 100\,\text{Вт}
  3. 50\,\text{Вт}
  4. 300\,\text{Вт}
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Given
m
20\,\text{кг}
h
6\,\text{м}
t
12\,\text{с}
g
10\,\text{Н/кг}
Find

P

  1. A=mghA=20 \cdot 10 \cdot 6A=1200\,\text{Дж}

    Lifting work increases potential energy.

  2. P=\frac{A}{t}P=\frac{1200}{12}P=100\,\text{Вт}

    Power is equal to work per unit time.

Answer100\,\text{Вт}

Position in this set: 13 from 27

Problem #157

Mechanical work and power

Load weight 30 kg evenly raised by 8 m for 20 s. Find the useful power at g = 10 N/kg.

Answer options

  1. 240\,\text{Вт}
  2. 60\,\text{Вт}
  3. 120\,\text{Вт}
  4. 360\,\text{Вт}
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Given
m
30\,\text{кг}
h
8\,\text{м}
t
20\,\text{с}
g
10\,\text{Н/кг}
Find

P

  1. A=mghA=30 \cdot 10 \cdot 8A=2400\,\text{Дж}

    Lifting work increases potential energy.

  2. P=\frac{A}{t}P=\frac{2400}{20}P=120\,\text{Вт}

    Power is equal to work per unit time.

Answer120\,\text{Вт}

Position in this set: 14 from 27

Problem #158

Mechanical work and power

Engine power 500 W works 20 s and transfers the cargo 0{,}6 proportion of energy expended. To what height will he lift a mass of 100 kg? g = 10 N/kg; the initial and final speeds are zero.

Answer options

  1. 12\,\text{м}
  2. 3\,\text{м}
  3. 6\,\text{м}
  4. 18\,\text{м}
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Given
P
500\,\text{Вт}
t
20\,\text{с}
r
0{,}6\,\text{доля}
m
100\,\text{кг}
g
10\,\text{Н/кг}
Find

h

  1. E=PtE=500 \cdot 20E=10000\,\text{Дж}

    We find the energy expended.

  2. A=\eta EA=0{,}6 \cdot 10000A=6000\,\text{Дж}

    We take into account the useful share of energy.

  3. h=\frac{A}{mg}h=\frac{6000}{(100 \cdot 10)}h=6\,\text{м}

    Useful work is equal to the increase in potential energy.

Answer6\,\text{м}

Position in this set: 15 from 27

Problem #159

Mechanical work and power

Engine power 800 W works 30 s and transfers the cargo 0{,}5 proportion of energy expended. To what height will he lift a mass of 150 kg? g = 10 N/kg; the initial and final speeds are zero.

Answer options

  1. 16\,\text{м}
  2. 4\,\text{м}
  3. 24\,\text{м}
  4. 8\,\text{м}
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Given
P
800\,\text{Вт}
t
30\,\text{с}
r
0{,}5\,\text{доля}
m
150\,\text{кг}
g
10\,\text{Н/кг}
Find

h

  1. E=PtE=800 \cdot 30E=24000\,\text{Дж}

    We find the energy expended.

  2. A=\eta EA=0{,}5 \cdot 24000A=12000\,\text{Дж}

    We take into account the useful share of energy.

  3. h=\frac{A}{mg}h=\frac{12000}{(150 \cdot 10)}h=8\,\text{м}

    Useful work is equal to the increase in potential energy.

Answer8\,\text{м}

Position in this set: 16 from 27

Problem #160

Levers and pulleys

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: 17 from 27

Problem #161

Levers and pulleys

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: 18 from 27

Problem #162

Levers and pulleys

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: 19 from 27

Problem #163

Levers and pulleys

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: 20 from 27

Problem #164

Levers and pulleys

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: 21 from 27

Problem #165

Levers and pulleys

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}

Position in this set: 22 from 27

Problem #166

Efficiency and mechanical energy

Useful installation work 300 J at energy expended 500 J. Find the efficiency as a percentage.

Answer options

  1. 120\,\text{\%}
  2. 30\,\text{\%}
  3. 60\,\text{\%}
  4. 180\,\text{\%}
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Given
A
300\,\text{Дж}
Q
500\,\text{Дж}
Find

\eta

  1. \eta=\frac{A}{Q}\cdot100\eta=\frac{300}{500} \cdot 100\eta=60\,\text{\%}

    Useful energy does not exceed expended energy.

Answer60\,\text{\%}

Position in this set: 23 from 27

Problem #167

Efficiency and mechanical energy

Useful installation work 450 J at energy expended 1000 J. Find the efficiency as a percentage.

Answer options

  1. 90\,\text{\%}
  2. 22{,}5\,\text{\%}
  3. 135\,\text{\%}
  4. 45\,\text{\%}
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Given
A
450\,\text{Дж}
Q
1000\,\text{Дж}
Find

\eta

  1. \eta=\frac{A}{Q}\cdot100\eta=\frac{450}{1000} \cdot 100\eta=45\,\text{\%}

    Useful energy does not exceed expended energy.

Answer45\,\text{\%}

Position in this set: 24 from 27

Problem #168

Efficiency and mechanical energy

Installation received 2000 J of energy and has efficiency 60 %. How much energy was wasted?

Answer options

  1. 1600\,\text{Дж}
  2. 400\,\text{Дж}
  3. 2400\,\text{Дж}
  4. 800\,\text{Дж}
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Given
Q
2000\,\text{Дж}
e
60\,\text{\%}
Find

R

  1. A=Q\frac{\eta}{100}A=\frac{2000 \cdot 60}{100}A=1200\,\text{Дж}

    We convert the percentage of efficiency into a share.

  2. R=Q-AR=2000 - 1200R=800\,\text{Дж}

    The losses account for the remaining energy.

Answer800\,\text{Дж}

Position in this set: 25 from 27

Problem #169

Efficiency and mechanical energy

Installation received 3000 J of energy and has efficiency 70 %. How much energy was wasted?

Answer options

  1. 900\,\text{Дж}
  2. 1800\,\text{Дж}
  3. 450\,\text{Дж}
  4. 2700\,\text{Дж}
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Given
Q
3000\,\text{Дж}
e
70\,\text{\%}
Find

R

  1. A=Q\frac{\eta}{100}A=\frac{3000 \cdot 70}{100}A=2100\,\text{Дж}

    We convert the percentage of efficiency into a share.

  2. R=Q-AR=3000 - 2100R=900\,\text{Дж}

    The losses account for the remaining energy.

Answer900\,\text{Дж}

Position in this set: 26 from 27

Problem #170

Efficiency and mechanical energy

Installation receives 10000 J. The first stage passes on 80 % of energy, the second turns into useful work 50 % of energy received by it. Find the total energy loss.

Answer options

  1. 6000\,\text{Дж}
  2. 12000\,\text{Дж}
  3. 3000\,\text{Дж}
  4. 18000\,\text{Дж}
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Given
Q
10000\,\text{Дж}
e_1
80\,\text{\%}
e_2
50\,\text{\%}
Find

R

  1. E=Q\frac{\eta_1}{100}E=\frac{10000 \cdot 80}{100}E=8000\,\text{Дж}

    In the second stage, only the energy after the first is available.

  2. A=E\frac{\eta_2}{100}A=\frac{8000 \cdot 50}{100}A=4000\,\text{Дж}

    The efficiencies of the stages are multiplied, not added.

  3. R=Q-AR=10000 - 4000R=6000\,\text{Дж}

    Subtract the final useful work from the initial energy.

Answer6000\,\text{Дж}

Position in this set: 27 from 27

Problem #171

Efficiency and mechanical energy

Installation receives 20000 J. The first stage passes on 75 % of energy, the second turns into useful work 60 % of energy received by it. Find the total energy loss.

Answer options

  1. 22000\,\text{Дж}
  2. 11000\,\text{Дж}
  3. 5500\,\text{Дж}
  4. 33000\,\text{Дж}
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Given
Q
20000\,\text{Дж}
e_1
75\,\text{\%}
e_2
60\,\text{\%}
Find

R

  1. E=Q\frac{\eta_1}{100}E=\frac{20000 \cdot 75}{100}E=15000\,\text{Дж}

    In the second stage, only the energy after the first is available.

  2. A=E\frac{\eta_2}{100}A=\frac{15000 \cdot 60}{100}A=9000\,\text{Дж}

    The efficiencies of the stages are multiplied, not added.

  3. R=Q-AR=20000 - 9000R=11000\,\text{Дж}

    Subtract the final useful work from the initial energy.

Answer11000\,\text{Дж}

The theory behind these problems

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

Mechanical work and powerWhen work is done and how we compare its rate.Levers and pulleysMoment arm, torque, lever equilibrium and the golden rule of mechanics.Efficiency and mechanical energyUseful and total input work, kinetic and potential energy.

Formulas by topic

Look up symbols and units of measurement in the reference.

Timet, τ · сEfficiencyη · %Torque (moment of a force)M · Н·мPowerP, N · ВтWorkA · ДжDistance and pathS, l · мEnergyE, W · Дж