Physics 1: Energy, Kinetic Energy, and Work

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Comprehensive practice flashcards covering kinetic energy, work, mechanical energy transformations, and problem-solving examples from Physics 1 lecture notes.

18 Terms

1
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According to the Law of Conservation of Energy (Energilagen), how does energy behave?

Energy can neither be created nor destroyed; it can only be converted or transferred.

2
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What units of measurement for energy are listed in the introductory lecture?

Joule (J\text{J}), Newton-meter (Nm\text{Nm}), Calorie (cal\text{cal}), and Kilowatt-hour (kWh\text{kWh}).

3
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What two primary forms constitute mechanical energy in the lecture?

Potential energy (lägesenergi) and kinetic energy (rörelseenergi).

4
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How is work (arbete) defined in relation to energy?

Work is defined as the transformation or conversion of energy (Arbete=omvandlingen\text{Arbete} = \text{omvandlingen}).

5
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What is the formula for kinetic energy (EkE_k), and what units do the variables use?

Ek=m⋅v22E_k = \frac{m \cdot v^2}{2}, where mm is mass in kilograms (kg\text{kg}), vv is velocity in meters per second (m/s\text{m/s}), and EkE_k is kinetic energy in Joules (J\text{J}).

6
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In the mini-lab, what is the kinetic energy of a golf ball with a mass of approximately 45 g45\,\text{g} rolling at 1.2 m/s1.2\,\text{m/s}?

Approximately 0.032 J0.032\,\text{J} (32 mJ32\,\text{mJ}), calculated as Ek=0.045 kg×(1.2 m/s)22≈0.0324 JE_k = \frac{0.045\,\text{kg} \times (1.2\,\text{m/s})^2}{2} \approx 0.0324\,\text{J}.

7
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If a tennis ball has a mass of 56 g56\,\text{g} and a kinetic energy of 0.69 J0.69\,\text{J}, what is its speed?

Approximately 5.0 m/s5.0\,\text{m/s} (≈4.96416 m/s\approx 4.96416\,\text{m/s}), calculated from v=2×0.69 J0.056 kgv = \sqrt{\frac{2 \times 0.69\,\text{J}}{0.056\,\text{kg}}}.

8
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For Kenny falling from a height of 8848 m8848\,\text{m} with a mass of 30 kg30\,\text{kg}, what is the initial potential energy (g=9.82 m/s2g = 9.82\,\text{m/s}^2)?

Ep=30 kg×9.82 m/s2×8848 m=2 606 620.8 JE_p = 30\,\text{kg} \times 9.82\,\text{m/s}^2 \times 8848\,\text{m} = 2\,606\,620.8\,\text{J}.

9
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Assuming all of Kenny's initial potential energy (2 606 620.8 J2\,606\,620.8\,\text{J}) converts to kinetic energy upon landing (m=30 kgm = 30\,\text{kg}), what is his calculated landing speed?

Approximately 420 m/s420\,\text{m/s} (≈416.863 m/s\approx 416.863\,\text{m/s}), calculated using v=2×2 606 620.8 J30 kgv = \sqrt{\frac{2 \times 2\,606\,620.8\,\text{J}}{30\,\text{kg}}}.

10
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How does Kenny's calculated landing speed of ≈420 m/s\approx 420\,\text{m/s} compare to the speed of sound given in the notes?

It is faster than the speed of sound, which is noted as vljud∼340 m/sv_{\text{ljud}} \sim 340\,\text{m/s}.

11
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What website and simulation are referenced for kinetic energy visualization?

phet.colorado.edu (energy skate park).

12
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What is the formula and standard units for mechanical work?

W=F⋅sW = F \cdot s, where WW is work in Joules (J\text{J}) or Newton-meters (Nm\text{Nm}), FF is force in Newtons (N\text{N}), and ss is displacement in meters (m\text{m}).

13
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What condition must be met regarding force and displacement for work to occur?

The force and the displacement must be in the same direction, and the object must actually be moved/affected.

14
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If a person applies a force of 500 N500\,\text{N} on a stone for 10 min10\,\text{min} but the stone does not move, how much work is performed on the stone?

0 J0\,\text{J}, because no displacement occurred (s=0 ms = 0\,\text{m}).

15
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How much work is required to lift a 65.0 kg65.0\,\text{kg} stone vertically onto a bench with a height of 72 cm72\,\text{cm} (g=9.82 m/s2g = 9.82\,\text{m/s}^2)?

Approximately 460 J460\,\text{J} (459.576 J459.576\,\text{J}), calculated as Fg=65.0 kg×9.82 m/s2≈638.3 NF_g = 65.0\,\text{kg} \times 9.82\,\text{m/s}^2 \approx 638.3\,\text{N} and W=638.3 N×0.72 mW = 638.3\,\text{N} \times 0.72\,\text{m}.

16
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How much work is needed to roll a stone up a 1.44 m1.44\,\text{m} long ramp at a 30∘30^\circ incline with a force of 319 N319\,\text{N}?

Approximately 459 J459\,\text{J} (459.36 J459.36\,\text{J}), calculated as W=319 N×1.44 mW = 319\,\text{N} \times 1.44\,\text{m}.

17
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If a bicycle has an initial kinetic energy of 1.6 kJ1.6\,\text{kJ} and stops with a braking force of 300 N300\,\text{N}, what is the braking distance?

Approximately 5.3 m5.3\,\text{m} (≈5.333 m\approx 5.333\,\text{m}), calculated using W=F⋅s⇒1600 J=300 N×sW = F \cdot s \Rightarrow 1600\,\text{J} = 300\,\text{N} \times s.

18
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When a cyclist applies the brakes to stop, what form of energy is the kinetic energy converted into?

Thermal energy / heat (värmeenergi).