GRADE 10 PRACTICE
Physics Problems: Conservation laws, Grade 10
Impulse, work, power and energy.
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.
Problem #4
Conservation laws: energy, work and powerSparrow mass 50\,\text{г} flies at altitude 20\,\text{м} at speed 12\,\text{м/с}. Determine the momentum and total mechanical energy of the sparrow.
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- m
- 50\,\text{г}=0{,}05\,\text{кг}
- h
- 20\,\text{м}
- v
- 12\,\text{м/с}
- g
- 10\,\tfrac{\text{м}}{\text{с}^{2}}
p - ?E_0 - ?
- p = mvp = 0{,}05\,\text{кг}\cdot 12\,\tfrac{\text{м}}{\text{с}}p = 0{,}6\,\tfrac{\text{кг}\cdot\text{м}}{\text{с}}
- E_0 = E_{\text{к}} + E_{\text{п}} = \dfrac{mv^{2}}{2} + mgh
Total mechanical energy consists of kinetic and potential.
- E_0 = \dfrac{mv^{2}}{2} + mghE_0 = \dfrac{0{,}05\,\text{кг}\cdot 12^{2}\,\tfrac{\text{м}^{2}}{\text{с}^{2}}}{2} + 0{,}05\,\text{кг}\cdot 10\,\tfrac{\text{м}}{\text{с}^{2}}\cdot 20\,\text{м}E_0 = 3{,}6\,\text{Дж} + 10\,\text{Дж} = 13{,}6\,\text{Дж}
Answerp = 0{,}6\,\tfrac{\text{кг}\cdot\text{м}}{\text{с}}E_0 = 13{,}6\,\text{Дж}
Problem #54
Conservation laws: energy, work and powerBody mass 400\,\text{г} begins to move under the influence of force 1{,}2\,\text{Н}. What speed will it acquire in 6\,\text{с}?
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- m
- 400\,\text{г}=0{,}4\,\text{кг}
- F
- 1{,}2\,\text{Н}
- t
- 6\,\text{с}
v - ?
- a = \dfrac{F}{m}a = \dfrac{1{,}2}{0{,}4}a = 3\,\tfrac{\text{м}}{\text{с}^{2}}
- v = atv = 3\cdot 6v = 18\,\tfrac{\text{м}}{\text{с}}
Through the force impulse it is shorter: Ft = mv, whence v = Ft/m.
Answerv = 18\,\tfrac{\text{м}}{\text{с}}
Problem #55
Conservation laws: energy, work and powerKite mass 1{,}5\,\text{кг} flies at speed 6\,\text{м/с}. Its total mechanical energy is 327\,\text{Дж}. Determine momentum, kinetic energy and altitude.
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- m
- 1{,}5\,\text{кг}
- v
- 6\,\text{м/с}
- E_0
- 327\,\text{Дж}
- g
- 10\,\tfrac{\text{м}}{\text{с}^{2}}
p - ?E_{\text{к}} - ?h - ?
- p = mvp = 1{,}5\cdot 6p = 9\,\tfrac{\text{кг}\cdot\text{м}}{\text{с}}
- E_{\text{к}} = \dfrac{mv^{2}}{2}E_{\text{к}} = \dfrac{1{,}5\cdot 6^{2}}{2}E_{\text{к}} = 27\,\text{Дж}
- E_0 = E_{\text{к}} + E_{\text{п}}\ \Rightarrow\ E_{\text{п}} = E_0 - E_{\text{к}}E_{\text{п}} = 327 - 27E_{\text{п}} = 300\,\text{Дж}
- E_{\text{п}} = mgh\ \Rightarrow\ h = \dfrac{E_{\text{п}}}{mg}h = \dfrac{300}{1{,}5\cdot 10}h = 20\,\text{м}
Answerp = 9\,\tfrac{\text{кг}\cdot\text{м}}{\text{с}}E_{\text{к}} = 27\,\text{Дж}h = 20\,\text{м}
Problem #56
Conservation laws: energy, work and powerboy mass 40\,\text{кг} rises from the first to the fifth floor in 2\,\text{мин}. Floor height 3\,\text{м}. How much power does it develop?
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- m
- 40\,\text{кг}
- h
- 12\,\text{м}
- t
- 2\,\text{мин}=120\,\text{с}
- g
- 10\,\tfrac{\text{м}}{\text{с}^{2}}
N - ?
- h = 4\cdot 3\,\text{м} = 12\,\text{м}
From the first floor to the fifth there are four flights, not five.
- N = \dfrac{A}{t} = \dfrac{mgh}{t}N = \dfrac{40\cdot 10\cdot 12}{120}N = 40\,\text{Вт}
AnswerN = 40\,\text{Вт}
Problem #57
Conservation Laws: MomentumMaterial point with mass 800\,\text{г} moves in a circle with radius 40\,\text{см} with frequency 5\,\text{об/с}. Determine the magnitude of the impulse. What is the modulus of change in momentum for a period, half and quarter of a period?
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- m
- 800\,\text{г}=0{,}8\,\text{кг}
- R
- 40\,\text{см}=0{,}4\,\text{м}
- \nu
- 5\,\text{Гц}
p - ?|\Delta\vec p_T| - ?|\Delta\vec p_{T/2}| - ?|\Delta\vec p_{T/4}| - ?
- v=2\pi R\nuv = 2\pi\cdot0{,}4\cdot5v = 12{,}57\,\text{м/с}
- p=mvp = 0{,}8\cdot12{,}57p = 10{,}05\,\text{кг}\cdot\text{м/с}
- |\Delta\vec p|=2p\sin\dfrac{\alpha}{2}|\Delta\vec p_T|=0,\quad |\Delta\vec p_{T/2}|=2p,\quad |\Delta\vec p_{T/4}|=\sqrt2p
Over the period, the vector returns to the original: the change is zero. In half the period it changes direction to the opposite; rotates 90° in a quarter.
Answerp=10{,}05\,\text{кг}\cdot\text{м/с}|\Delta\vec p_T|=0\,\text{кг}\cdot\text{м/с}|\Delta\vec p_{T/2}|=20{,}11\,\text{кг}\cdot\text{м/с}|\Delta\vec p_{T/4}|=14{,}22\,\text{кг}\cdot\text{м/с}
Problem #58
Conservation Laws: MomentumBullet mass 10\,\text{г} travels horizontally at a speed of 800\,\text{м/с}, passes through an initially stationary sphere of mass 2\,\text{кг} suspended from a long massless string and emerges in its original direction at a speed of 300\,\text{м/с}. How high above its initial position will the sphere rise? Treat the impact as brief and neglect energy losses during the subsequent rise.
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- m
- 10\,\text{г}=0{,}01\,\text{кг}
- v_0
- 800\,\text{м/с}
- v
- 300\,\text{м/с}
- M
- 2\,\text{кг}
- g
- 10\,\tfrac{\text{м}}{\text{с}^{2}}
h - ?
- mv_0 = Mu + mv\ \Rightarrow\ Mu = m\left(v_0 - v\right)
The law of conservation of momentum when breaking through a ball.
- u = \dfrac{m\left(v_0 - v\right)}{M}u = \dfrac{0{,}01\left(800 - 300\right)}{2}u = 2{,}5\,\tfrac{\text{м}}{\text{с}}
- \dfrac{Mu^{2}}{2} = Mgh\ \Rightarrow\ h = \dfrac{u^{2}}{2g}h = \dfrac{2{,}5^{2}}{2\cdot 10}h = 0{,}31\,\text{м}
Answerh = 0{,}31\,\text{м}
Problem #59
Conservation laws: energy, work and powerA boy carries a sled, applying it to the rope at an angle 30^{\circ} force 8\,\text{Н}. For 15\,\text{с} the sled passed 30\,\text{м}. Define work and power.
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- \alpha
- 30^{\circ}
- F
- 8\,\text{Н}
- t
- 15\,\text{с}
- S
- 30\,\text{м}
A - ?N - ?
- A = F S\cos\alphaA = 8\cdot 30\cdot 0{,}866A = 208\,\text{Дж}
Work is performed only by the component of the force along the displacement.
- N = \dfrac{A}{t}N = \dfrac{208\,\text{Дж}}{15\,\text{с}}N = 13{,}9\,\text{Вт}
AnswerA = 208\,\text{Дж}N = 13{,}9\,\text{Вт}
Problem #60
Conservation laws: energy, work and powerUnder force 34\,\text{Н} body mass 2\,\text{кг} rose to heights 15\,\text{м} from a state of rest. Determine the kinetic energy at this height.
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- F
- 34\,\text{Н}
- m
- 2\,\text{кг}
- h
- 15\,\text{м}
- v_0
- 0
E_{\text{к}} - ?
- A = \Delta E_0\ \Rightarrow\ Fh = mgh + E_{\text{к}}
The work of force goes both for lifting and acceleration.
- E_{\text{к}} = Fh - mgh34\cdot 15 - 2\cdot 10\cdot 15 = 510 - 300E_{\text{к}} = 210\,\text{Дж}
AnswerE_{\text{к}} = 210\,\text{Дж}
Problem #61
Conservation laws: energy, work and powerBody mass 2\,\text{кг} falls from a height 8\,\text{м}. Determine the kinetic energy at altitude 3\,\text{м} above the Earth.
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- m
- 2\,\text{кг}
- h
- 8\,\text{м}
- h_2
- 3\,\text{м}
- g
- 10\,\tfrac{\text{м}}{\text{с}^{2}}
E_{\text{к}2} - ?
- mgh + E_{\text{к}0} = mgh_2 + E_{\text{к}2},\quad E_{\text{к}0} = 0
The total mechanical energy is conserved.
- E_{\text{к}2} = mg\left(h - h_2\right)2\cdot 10\left(8 - 3\right)E_{\text{к}2} = 100\,\text{Дж}
AnswerE_{\text{к}2} = 100\,\text{Дж}
Problem #63
Conservation Laws: MomentumIcebreaker mass 1000\,\text{т} collided with a stationary iceberg at speed 0{,}5\,\text{м/с}. After the collision, the speed of the icebreaker became 0{,}2\,\text{м/с}. Determine the mass of the iceberg, assuming joint motion.
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- m_1
- 1000\,\text{т}
- v_1
- 0{,}5\,\text{м/с}
- v
- 0{,}2\,\text{м/с}
m_2 - ?
- m_1 v_1 = \left(m_1 + m_2\right)v
The collision is inelastic: the bodies then move together.
- m_1 + m_2 = \dfrac{m_1 v_1}{v}m_2 = \dfrac{1000\cdot 0{,}5}{0{,}2}m_2 = 2500\,\text{т}
- m_2 = \left(m_1 + m_2\right) - m_1m_2 = 2500 - 1000m_2 = 1500\,\text{т}
Answerm_2 = 1500\,\text{т}
Problem #64
Conservation laws: energy, work and powerBullet mass 10\,\text{г} has kinetic energy 3200\,\text{Дж}. Determine the recoil velocity of a rifle with a mass 4\,\text{кг} and the work of powder gases.
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- m_1
- 10\,\text{г}=0{,}01\,\text{кг}
- E_{\text{к}1}
- 3200\,\text{Дж}
- m_2
- 4\,\text{кг}
v_2 - ?A - ?
- E_{\text{к}1} = \dfrac{m_1 v_1^{2}}{2}\ \Rightarrow\ v_1 = \sqrt{\dfrac{2E_{\text{к}1}}{m_1}}v_1 = \sqrt{\dfrac{2\cdot 3200}{0{,}01}}v_1 = 800\,\tfrac{\text{м}}{\text{с}}
- m_1 v_1 = m_2 v_2\ \Rightarrow\ v_2 = \dfrac{m_1 v_1}{m_2}v_2 = \dfrac{0{,}01\cdot 800}{4}v_2 = 2\,\tfrac{\text{м}}{\text{с}}
- A = E_{\text{к}1} + \dfrac{m_2 v_2^{2}}{2}A = 3200 + \dfrac{4\cdot 2^{2}}{2}A = 3208\,\text{Дж}
Answerv_2 = 2\,\tfrac{\text{м}}{\text{с}}A = 3208\,\text{Дж}
Problem #65
Conservation laws: energy, work and powerCar weight 20\,\text{т} moves at speed 0{,}5\,\text{м/с} towards the car with a mass 30\,\text{т} at speed 0{,}4\,\text{м/с}. Determine the total velocity after the inelastic collision and the amount of heat generated by the collision. Select the positive direction of the axis based on the initial speed of the first car.
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- m_1
- 20\,\text{т}
- v_1
- 0{,}5\,\text{м/с}
- m_2
- 30\,\text{т}
- v_2
- 0{,}4\,\text{м/с}
v_x - ?Q - ?
- m_1 v_1 - m_2 v_2 = \left(m_1 + m_2\right)v_x
The cars are moving towards each other, so the impulses are subtracted.
- v_x = \dfrac{m_1 v_1 - m_2 v_2}{m_1 + m_2}\dfrac{20\cdot 0{,}5 - 30\cdot 0{,}4}{20 + 30} = \dfrac{10 - 12}{50}v_x = -0{,}04\,\tfrac{\text{м}}{\text{с}}
A minus sign means that the coupler will move towards the second car.
- Q = \dfrac{m_1 v_1^{2}}{2} + \dfrac{m_2 v_2^{2}}{2} - \dfrac{\left(m_1 + m_2\right)v_x^{2}}{2}Q = \dfrac{20\cdot 0{,}5^{2}}{2} + \dfrac{30\cdot 0{,}4^{2}}{2} - \dfrac{50\cdot 0{,}04^{2}}{2}Q = 4{,}86\,\text{кДж} = 4860\,\text{Дж}
Answerv_x = -0{,}04\,\tfrac{\text{м}}{\text{с}}Q = 4860\,\text{Дж}
Problem #66
Conservation laws: energy, work and powerVehicle weight 800\,\text{кг} moves at speed 54\,\text{км/ч}. The friction coefficient is equal to 0{,}08. How much power does the engine develop and how much work does it do per minute with uniform motion?
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- m
- 800\,\text{кг}
- \mu
- 0{,}08
- v
- 54\,\text{км/ч}=15\,\text{м/с}
- t
- 1\,\text{мин}=60\,\text{с}
N - ?A - ?
- F_{\text{т}} = F_{\text{тр}} = \mu m gF_{\text{т}} = 0{,}08\cdot 800\cdot 10F_{\text{т}} = 640\,\text{Н}
At constant speed, the thrust is equal to the friction force.
- N = F_{\text{т}}\,vN = 640\cdot 15N = 9600\,\text{Вт} = 9{,}6\,\text{кВт}
- A = N tA = 9600\cdot 60A = 576\,000\,\text{Дж}
AnswerN = 9600\,\text{Вт}A = 576\,000\,\text{Дж}
Problem #67
Conservation laws: energy, work and powerIn a spring gun, the spring is compressed by 5\,\text{см}. Determine the spring stiffness if the bullet weighs 8\,\text{г} flies out at speed 5\,\text{м/с}.
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- \Delta\ell
- 5\,\text{см}=0{,}05\,\text{м}
- m
- 8\,\text{г}=0{,}008\,\text{кг}
- v
- 5\,\text{м/с}
k - ?
- \dfrac{k\,\Delta\ell^{2}}{2} = \dfrac{mv^{2}}{2}
The energy of the compressed spring is completely converted into the kinetic energy of the bullet.
- k = \dfrac{mv^{2}}{\Delta\ell^{2}}k = \dfrac{0{,}008\cdot 5^{2}}{0{,}05^{2}}k = 80\,\tfrac{\text{Н}}{\text{м}}
Answerk = 80\,\tfrac{\text{Н}}{\text{м}}
Problem #68
Conservation laws: energy, work and powerArrow mass 40\,\text{г} launched vertically upward at a speed 50\,\text{м/с}. Determine its kinetic energy at height 100\,\text{м}.
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- m
- 40\,\text{г}=0{,}04\,\text{кг}
- v_0
- 50\,\text{м/с}
- h
- 100\,\text{м}
- g
- 10\,\tfrac{\text{м}}{\text{с}^{2}}
E_{\text{к}} - ?
- \dfrac{mv_0^{2}}{2} = mgh + E_{\text{к}}\ \Rightarrow\ E_{\text{к}} = \dfrac{mv_0^{2}}{2} - mgh
Part of the initial kinetic energy turned into potential.
- E_{\text{к}} = \dfrac{mv_0^{2}}{2} - mgh\dfrac{0{,}04\cdot 50^{2}}{2} - 0{,}04\cdot 10\cdot 100 = 50 - 40E_{\text{к}} = 10\,\text{Дж}
AnswerE_{\text{к}} = 10\,\text{Дж}
Problem #69
Conservation laws: energy, work and powerThe recoil speed of the gun barrel when fired is equal to 20\,\text{м/с}. Barrel weight 800\,\text{кг}, projectile mass 16\,\text{кг}. Determine the speed of the projectile.
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- v_1
- 20\,\text{м/с}
- m_1
- 800\,\text{кг}
- m_2
- 16\,\text{кг}
v_2 - ?
- m_1 v_1 = m_2 v_2\ \Rightarrow\ v_2 = \dfrac{m_1 v_1}{m_2}v_2 = \dfrac{800\cdot 20}{16}v_2 = 1000\,\tfrac{\text{м}}{\text{с}}
Law of conservation of momentum: the total momentum before and after is zero.
Answerv_2 = 1000\,\tfrac{\text{м}}{\text{с}}
Problem #70
Conservation laws: energy, work and powerBody mass 5\,\text{кг} when falling to Earth it has kinetic energy 160\,\text{Дж}. From what height did it fall?
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- m
- 5\,\text{кг}
- E_{\text{к}}
- 160\,\text{Дж}
- g
- 10\,\tfrac{\text{м}}{\text{с}^{2}}
h - ?
- mgh = E_{\text{к}}\ \Rightarrow\ h = \dfrac{E_{\text{к}}}{mg}h = \dfrac{160}{5\cdot 10}h = 3{,}2\,\text{м}
During a fall from rest, potential energy is converted into kinetic energy; air resistance is neglected.
Answerh = 3{,}2\,\text{м}
Problem #159
Momentum and its conservationObject mass 3 kg, speed 5 m/s. Find the modulus of the impulse.
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- m
- 3\,\text{кг}
- v
- 5\,\text{м/с}
p - ?
- p=mvp=3 \cdot 5p=15\,\text{кг·м/с}
The impulse is directed in the same way as the speed.
Answer15\,\text{кг·м/с}
Problem #160
Momentum and its conservationObject mass 4 kg, speed 7 m/s. Find the modulus of the impulse.
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- m
- 4\,\text{кг}
- v
- 7\,\text{м/с}
p - ?
- p=mvp=4 \cdot 7p=28\,\text{кг·м/с}
The impulse is directed in the same way as the speed.
Answer28\,\text{кг·м/с}
Problem #161
Momentum and its conservationTrolley weight 2 kg moves with speed 6 m/s and engages with a stationary trolley mass 4 kg. Find the overall speed. Neglect external impulse.
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- m_1
- 2\,\text{кг}
- m_2
- 4\,\text{кг}
- v
- 6\,\text{м/с}
u - ?
- p=m_1vp=2 \cdot 6p=12\,\text{кг·м/с}
Before the impact, only the first cart moves.
- u=\frac{p}{m_1+m_2}u=\frac{12}{(2 + 4)}u=2\,\text{м/с}
After coupling, the momentum belongs to the total mass.
Answer2\,\text{м/с}
Problem #162
Momentum and its conservationTrolley weight 3 kg moves with speed 8 m/s and engages with a stationary trolley mass 3 kg. Find the overall speed. Neglect external impulse.
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- m_1
- 3\,\text{кг}
- m_2
- 3\,\text{кг}
- v
- 8\,\text{м/с}
u - ?
- p=m_1vp=3 \cdot 8p=24\,\text{кг·м/с}
Before the impact, only the first cart moves.
- u=\frac{p}{m_1+m_2}u=\frac{24}{(3 + 3)}u=4\,\text{м/с}
After coupling, the momentum belongs to the total mass.
Answer4\,\text{м/с}
Problem #163
Momentum and its conservationCarts by the masses 4 and 2 kg are moving towards at speeds 5 and 4 m/s and interlock. Find the modulus of the overall speed. The positive direction coincides with the movement of the first cart.
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- m_1
- 4\,\text{кг}
- m_2
- 2\,\text{кг}
- v_1
- 5\,\text{м/с}
- v_2
- 4\,\text{м/с}
u - ?
- p_1=m_1v_1p_1=4 \cdot 5p_1=20\,\text{кг·м/с}
The first impulse is positive.
- |p_2|=m_2v_2|p_2|=2 \cdot 4|p_2|=8\,\text{кг·м/с}
The second impulse is directed opposite to the first.
- u=\frac{p_1-|p_2|}{m_1+m_2}u=\frac{(20 - 8)}{(4 + 2)}u=2\,\text{м/с}
In these data, the final impulse is directed along the movement of the first cart.
Answer2\,\text{м/с}
Problem #164
Momentum and its conservationCarts by the masses 5 and 3 kg are moving towards at speeds 6 and 2 m/s and interlock. Find the modulus of the overall speed. The positive direction coincides with the movement of the first cart.
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- m_1
- 5\,\text{кг}
- m_2
- 3\,\text{кг}
- v_1
- 6\,\text{м/с}
- v_2
- 2\,\text{м/с}
u - ?
- p_1=m_1v_1p_1=5 \cdot 6p_1=30\,\text{кг·м/с}
The first impulse is positive.
- |p_2|=m_2v_2|p_2|=3 \cdot 2|p_2|=6\,\text{кг·м/с}
The second impulse is directed opposite to the first.
- u=\frac{p_1-|p_2|}{m_1+m_2}u=\frac{(30 - 6)}{(5 + 3)}u=3\,\text{м/с}
In these data, the final impulse is directed along the movement of the first cart.
Answer3\,\text{м/с}
Problem #165
Work, power and 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 #166
Work, power and 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 #167
Work, power and 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 #168
Work, power and 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 #169
Work, power and 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 #170
Work, power and 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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