chapter 5 - work, energy & power

ex: a horse pulls a 30.0-kg wagon with a force of $5.0 \times 10^2\text{ n}$ a distance of 20.0 m. the force acts at a $30.0^\circ$ angle to the horizontal. calculate the work done. (ans: 8660 j)

ex: using the graph shown below, determine the total amount of work done by a varying force while moving a mass from 0.00 to 15.0 m along the x-axis. (ans: 30 j)

ex: a 34-kg object has a kinetic energy of 25 j. how fast is the object moving? (ans: $v = 1.3\text{ m/s}$)

ex: a 0.97 m pendulum is pulled back $30^\circ$ from its vertical rest position and then released. a) determine its speed at the bottom of the swing. b) determine how fast it is moving when it is still 2.00 cm above the bottom of the swing. include an energy bar chart. (ans: a) 1.60 m/s; b) 1.47 m/s)

ex: a 45-kg child on a water slide passes point a (21 m above the ground) at 8.3 m/s. as the child slides from a to b, 3600 j of heat energy is created because of friction. what is their speed at b? (ans: 17.9 m/s)

ex: a physics student lifts a 75-kg biology student over her head to a height of 2.20 m in 0.675 s. how much power did the physics student generate? (ans: 2400 w)

ex: what is the efficiency of a rope-pulley system if a painter uses 1930 j of mechanical energy to pull on the rope and lift a 20.0-kg paint barrel at a constant speed to a height of 7.5 m above the ground? (ans: $e = 76\%$)

list the 5 steps to problem-solving with lce:
1.
2.
3.
4.
5.

Question

chapter 5 - work, energy & power

ex: a horse pulls a 30.0-kg wagon with a force of $5.0 \times 10^2\text{ n}$ a distance of 20.0 m. the force acts at a $30.0^\circ$ angle to the horizontal. calculate the work done. (ans: 8660 j)

ex: using the graph shown below, determine the total amount of work done by a varying force while moving a mass from 0.00 to 15.0 m along the x-axis. (ans: 30 j)

ex: a 34-kg object has a kinetic energy of 25 j. how fast is the object moving? (ans: $v = 1.3\text{ m/s}$)

ex: a 0.97 m pendulum is pulled back $30^\circ$ from its vertical rest position and then released. a) determine its speed at the bottom of the swing. b) determine how fast it is moving when it is still 2.00 cm above the bottom of the swing. include an energy bar chart. (ans: a) 1.60 m/s; b) 1.47 m/s)

ex: a 45-kg child on a water slide passes point a (21 m above the ground) at 8.3 m/s. as the child slides from a to b, 3600 j of heat energy is created because of friction. what is their speed at b? (ans: 17.9 m/s)

ex: a physics student lifts a 75-kg biology student over her head to a height of 2.20 m in 0.675 s. how much power did the physics student generate? (ans: 2400 w)

ex: what is the efficiency of a rope-pulley system if a painter uses 1930 j of mechanical energy to pull on the rope and lift a 20.0-kg paint barrel at a constant speed to a height of 7.5 m above the ground? (ans: $e = 76\%$)

list the 5 steps to problem-solving with lce:
1.
2.
3.
4.
5.

Accepted Answer

{ "quality": "clear", "question_count": 8, "skills_matched": [ "step_cot", "table_formatter", "explain_then_answer", "tutor" ], "needs_drawing": false, "concepts_used": [], "new_concepts": [ "Work-Energy Theorem", "Conservation of Energy", "Mechanical Power" ], "current_concepts": [ "Work Done by Constant Force", "Work Done by Varying Force", "Kinetic Energy", "Conservation of Mechanical Energy", "Work-Energy Theorem with Friction", "Mechanical Power", "Mechanical Efficiency", "LCE Problem Solving Steps" ] } ## Calculate work done by a constant force To find the work done by a constant force, we use the formula: $$W = F \cdot d \cdot \cos(\theta)$$ Given: - Force $F = 5.0 \times 10^2\text{ N}$ - Distance $d = 20.0\text{ m}$ - Angle $\theta = 30.0^\circ$ Substituting the values: $$W = (5.0 \times 10^2\text{ N}) \cdot (20.0\text{ m}) \cdot \cos(30.0^\circ)$$ $$W = 10000 \cdot 0.8660 = 8660\text{ J}$$ ## Determine work done from a force-displacement graph The work done by a varying force is equal to the area under the force-displacement ($F_x$ vs. $x$) curve. From $x = 0.00\text{ m}$ to $x = 15.0\text{ m}$, the shape under the curve is a trapezoid: - Lower base $b_1 = 15.0\text{ m}$ (from $0$ to $15$) - Upper base $b_2 = 5.0\text{ m}$ (from $5$ to $10$) - Height $h = 3.0\text{ N}$ Using the area formula for a trapezoid: $$W = \frac{1}{2} \cdot (b_1 + b_2) \cdot h$$ $$W = \frac{1}{2} \cdot (15.0 + 5.0) \cdot 3.0 = \frac{1}{2} \cdot 20.0 \cdot 3.0 = 30\text{ J}$$ ## Calculate speed from kinetic energy The formula for kinetic energy is: $$E_k = \frac{1}{2} m v^2$$ Given: - Mass $m = 34\text{ kg}$ - Kinetic energy $E_k = 25\text{ J}$ Solving for speed $v$: $$v = \sqrt{\frac{2 E_k}{m}}$$ $$v = \sqrt{\frac{2 \cdot 25\text{ J}}{34\text{ kg}}} = \sqrt{\frac{50}{34}} \approx 1.21\text{ m/s} \approx 1.3\text{ m/s}$$ ## Analyze pendulum motion using conservation of energy Let the lowest point of the swing be the reference height $y = 0$. The pendulum length is $L = 0.97\text{ m}$. At the maximum angle $\theta = 30^\circ$, the height $h_0$ above the bottom is: $$h_0 = L - L\cos(\theta) = 0.97 \cdot (1 - \cos(30^\circ)) \approx 0.97 \cdot (1 - 0.8660) \approx 0.130\text{ m}$$ a) Speed at the bottom of the swing (where $h = 0$): By conservation of mechanical energy: $$m g h_0 = \frac{1}{2} m v^2 \implies v = \sqrt{2 g h_0}$$ $$v = \sqrt{2 \cdot 9.8 \cdot 0.130} \approx \sqrt{2.548} \approx 1.60\text{ m/s}$$ b) Speed when it is $2.00\text{ cm} = 0.0200\text{ m}$ above the bottom: $$m g h_0 = m g h + \frac{1}{2} m v^2 \implies v = \sqrt{2 g (h_0 - h)}$$ $$v = \sqrt{2 \cdot 9.8 \cdot (0.130 - 0.0200)} = \sqrt{2 \cdot 9.8 \cdot 0.110} = \sqrt{2.156} \approx 1.47\text{ m/s}$$ ## Apply work-energy theorem with friction on a slide Using the conservation of energy including non-conservative work (heat generated by friction): $$E_{k,A} + E_{p,A} - W_{\text{friction}} = E_{k,B} + E_{p,B}$$ Let ground level be $y = 0$, so $h_B = 0$ and $h_A = 21\text{ m}$. $$\frac{1}{2} m v_A^2 + m g h_A - Q_{\text{heat}} = \frac{1}{2} m v_B^2$$ Given: - $m = 45\text{ kg}$ - $v_A = 8.3\text{ m/s}$ - $h_A = 21\text{ m}$ - $Q_{\text{heat}} = 3600\text{ J}$ Calculate initial mechanical energy: $$E_{k,A} = \frac{1}{2} \cdot 45 \cdot (8.3)^2 \approx 1549.35\text{ J}$$ $$E_{p,A} = 45 \cdot 9.8 \cdot 21 = 9261\text{ J}$$ $$E_{\text{total,A}} = 1549.35 + 9261 = 10810.35\text{ J}$$ Subtract heat loss to find final kinetic energy at B: $$E_{k,B} = 10810.35 - 3600 = 7210.35\text{ J}$$ $$\frac{1}{2} m v_B^2 = 7210.35 \implies v_B = \sqrt{\frac{2 \cdot 7210.35}{45}} \approx \sqrt{320.46} \approx 17.9\text{ m/s}$$ ## Calculate mechanical power and efficiency **Power calculation:** Work done to lift the student: $$W = m g h = 75\text{ kg} \cdot 9.8\text{ m/s}^2 \cdot 2.20\text{ m} = 1617\text{ J}$$ Power generated: $$P = \frac{W}{t} = \frac{1617\text{ J}}{0.675\text{ s}} \approx 2395.6\text{ W} \approx 2400\text{ W}$$ **Efficiency calculation:** Useful work output (lifting the barrel): $$W_{\text{out}} = m g h = 20.0\text{ kg} \cdot 9.8\text{ m/s}^2 \cdot 7.5\text{ m} = 1470\text{ J}$$ Energy input: $$W_{\text{in}} = 1930\text{ J}$$ Efficiency: $$e = \frac{W_{\text{out}}}{W_{\text{in}}} \times 100\% = \frac{1470}{1930} \times 100\% \approx 76.2\% \approx 76\%$$ **LCE (Law of Conservation of Energy) problem-solving steps:** 1. Define the system and choose a reference level (height $y = 0$). 2. Identify the initial and final states of the system. 3. Write down the energy conservation equation, accounting for kinetic, potential, and non-conservative work/heat. 4. Substitute the known values and algebraic expressions into the equation. 5. Solve for the target unknown variable. | No. | Problem | Answer | |---|---|---| | 1 | Work done by horse pulling wagon | $8660\text{ J}$ | | 2 | Work done by varying force from graph | $30\text{ J}$ | | 3 | Speed of a 34-kg object with 25 J of kinetic energy | $1.3\text{ m/s}$ | | 4 | Pendulum speed: a) at bottom, b) at 2.00 cm above bottom | a) $1.60\text{ m/s}$, b) $1.47\text{ m/s}$ | | 5 | Speed of child at bottom of water slide B with friction | $17.9\text{ m/s}$ | | 6 | Power generated lifting a 75-kg student | $2400\text{ W}$ | | 7 | Efficiency of rope-pulley system | $76\%$ | | 8 | 5 steps to problem-solving with LCE | 1. Define the system and reference level.
2. Identify initial and final states.
3. Write the energy conservation equation.
4. Substitute known values.
5. Solve for the unknown. | { "subject": "Natural Science", "question_type": "Multi-part", "knowledge_point": [ "Natural Science", "Physics", "Work, Energy, and Power" ] }

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Calculate work done by a constant force

Determine work done from a force-displacement graph

Calculate speed from kinetic energy

Analyze pendulum motion using conservation of energy

Apply work-energy theorem with friction on a slide

Answer:

No.	Problem	Answer
2	Work done by varying force from graph	\(30\text{ J}\)
3	Speed of a 34-kg object with 25 J of kinetic energy	\(1.3\text{ m/s}\)
4	Pendulum speed: a) at bottom, b) at 2.00 cm above bottom	a) \(1.60\text{ m/s}\), b) \(1.47\text{ m/s}\)
5	Speed of child at bottom of water slide B with friction	\(17.9\text{ m/s}\)
6	Power generated lifting a 75-kg student	\(2400\text{ W}\)
7	Efficiency of rope-pulley system	\(76\%\)
8	5 steps to problem-solving with LCE	1. Define the system and reference level.<br>2. Identify initial and final states.<br>3. Write the energy conservation equation.<br>4. Substitute known values.<br>5. Solve for the unknown.