GRADE 9 PRACTICE
Physics Problems: Work and Energy, Grade 9
Work, power and conservation of mechanical energy.
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Problem statements
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Problem #25
Work, power and conservation of energyWhat concept does the description correspond to: “The sum of kinetic and potential energy describes the ability of a system to perform mechanical work.”?
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- \text{Description}
- \text{The sum of kinetic and potential energy describes the ability of a system to do mechanical work.}
\text{name concepts} - ?
- \text{Let's compare approval with definition}
Kinetic energy depends on speed; potential energy depends on position or deformation.
- \text{Let's formulate conclusion}\text{Concept} = \text{Mechanical energy}
Answer\text{Concept} = \text{Mechanical energy}
Problem #26
Work, power and conservation of energyWhat concept does the description correspond to: “The work of a constant force depends on its module, displacement and the angle between them.”?
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- \text{Description}
- \text{Work done by a constant force depends on its magnitude, the displacement and the angle between them.}
\text{name concepts} - ?
- \text{Let's compare approval with definition}
A force perpendicular to velocity changes the direction of motion but does no mechanical work at that instant.
- \text{Let's formulate conclusion}\text{Concept} = \text{Work done by a force}
Answer\text{Concept} = \text{Work done by a force}
Problem #27
EnergyA ball of mass 2 kg moves at a speed of 6 m/s. Find its kinetic energy.
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- m
- 2\,\text{кг}
- v
- 6\,\text{м/с}
E_k - ?
- E_k=\frac{mv^2}{2}E_k=\frac{2\cdot6^2}{2}\,\text{Дж}E_k = 36\,\text{Дж}
Answer36\,\text{Дж}
Problem #112
Work, power and conservation of energyForce 40 H is directed along the movement 5 m. Find the work done by the force.
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- F
- 40\,\text{Н}
- S
- 5\,\text{м}
A - ?
- A=FSA=40 \cdot 5A=200\,\text{Дж}
Force and displacement are co-directed.
Answer200\,\text{Дж}
Problem #113
Work, power and conservation of energyForce 60 H is directed along the movement 8 m. Find the work done by the force.
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- F
- 60\,\text{Н}
- S
- 8\,\text{м}
A - ?
- A=FSA=60 \cdot 8A=480\,\text{Дж}
Force and displacement are co-directed.
Answer480\,\text{Дж}
Problem #114
Work, power and conservation of energyLoad weight 20 kg evenly raised by 6 m for 12 s. Find the useful power at g = 10 N/kg.
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- m
- 20\,\text{кг}
- h
- 6\,\text{м}
- t
- 12\,\text{с}
- g
- 10\,\text{Н/кг}
P - ?
- A=mghA=20 \cdot 10 \cdot 6A=1200\,\text{Дж}
Lifting work increases potential energy.
- P=\frac{A}{t}P=\frac{1200}{12}P=100\,\text{Вт}
Power is equal to work per unit time.
Answer100\,\text{Вт}
Problem #115
Work, power and conservation of energyLoad weight 30 kg evenly raised by 8 m for 20 s. Find the useful power at g = 10 N/kg.
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- m
- 30\,\text{кг}
- h
- 8\,\text{м}
- t
- 20\,\text{с}
- g
- 10\,\text{Н/кг}
P - ?
- A=mghA=30 \cdot 10 \cdot 8A=2400\,\text{Дж}
Lifting work increases potential energy.
- P=\frac{A}{t}P=\frac{2400}{20}P=120\,\text{Вт}
Power is equal to work per unit time.
Answer120\,\text{Вт}
Problem #116
Work, power and conservation of energyEngine 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.
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- P
- 500\,\text{Вт}
- t
- 20\,\text{с}
- r
- 0{,}6\,\text{доля}
- m
- 100\,\text{кг}
- g
- 10\,\text{Н/кг}
h - ?
- E=PtE=500 \cdot 20E=10000\,\text{Дж}
We find the energy expended.
- A=\eta EA=0{,}6 \cdot 10000A=6000\,\text{Дж}
We take into account the useful share of energy.
- 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}
Problem #117
Work, power and conservation of energyEngine 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.
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- P
- 800\,\text{Вт}
- t
- 30\,\text{с}
- r
- 0{,}5\,\text{доля}
- m
- 150\,\text{кг}
- g
- 10\,\text{Н/кг}
h - ?
- E=PtE=800 \cdot 30E=24000\,\text{Дж}
We find the energy expended.
- A=\eta EA=0{,}5 \cdot 24000A=12000\,\text{Дж}
We take into account the useful share of energy.
- 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}
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