Generate a test of work power energy 10 question

To answer any work–power–energy question, keep three key formulas ready:\n\n1. Work ($W$): (W = \vec F \cdot \vec s = F s \cos \theta) when force is constant.\n2. Kinetic Energy (\text{KE}): (\text{KE} = \frac{1}{2} m v^2).\n3. Potential Energy (\text{PE}): (\text{PE}{\text{grav}} = m g h) and (\text{PE}{\text{spring}} = \frac{1}{2} k x^2).\n4. Power (P): (P = \frac{W}{t}) or (P = \vec F \cdot \vec v).\n\nThe usual logic a student follows:\n\n* Identify what is asked: work, energy, power, speed, height, etc.\n* Note the given data: mass, speed, force, time, distance.\n* Choose the right formula from the four above.\n* Substitute numbers carefully with correct units.\n* If more than one energy changes, apply conservation of mechanical energy: initial total energy = final total energy (when non-conservative work is zero).\n* For variable forces, integrate: \n $\n W = \int F(x)\,dx\n$\n* Always keep an eye on units: 1 kW = 1000 W, 72 km/h = 20 m/s, etc.

Imagine pushing your toy car up a ramp.\n\n* Work is how hard you push and how far the car moves.\n* Energy is like the fuel that lets the car move or helps you push.\n* Power is how fast you can finish pushing.\n\nThese 10 questions will check if you understand these ideas and can do simple maths with them.

10-Question Test (Basic Indian English)\n\nQ1. A constant force of 10 N moves a box 2 m in same line of force. Find work done.\n\nQ2. A 0.5 kg ball is thrown straight up with speed 20 m/s. Using energy method, find the highest point it reaches above the launch place.\n\nQ3. A 2 kg object moves at 5 m/s. What is its kinetic energy?\n\nQ4. A spring of constant 200 N/m is compressed by 0.1 m. Find energy stored in the spring.\n\nQ5. A motor pump lifts 100 kg water to 20 m height in 10 s. Find average power of motor in watt.\n\nQ6. A car (1000 kg) travels at 72 km/h and stops by applying brakes. How much work is done by brakes?\n\nQ7. A force (F(x)=3x^2,\text{N}) acts along x-axis. Find work when object moves from x = 0 m to x = 2 m.\n\nQ8. A machine delivers power 5 kW. How much work will it do in 3 minutes?\n\nQ9. If net work done on a body is zero, what happens to its kinetic energy?\n\nQ10. A 1 kg pendulum bob on a 2 m string is released from 60° with vertical. Find its speed at lowest point (ignore air drag).\n\n---\n\n#### Answer Key with Steps\n\nA1. (W = F s = 10 \times 2 = 20,\text{J}).\n\nA2. Initial (\text{KE} = \frac{1}{2}\times0.5\times20^2 = 100,\text{J}).\n\n Use (m g h = \text{KE}\Rightarrow h = \frac{100}{0.5\times9.8} \approx 20.4,\text{m}).\n\nA3. (\text{KE} = \frac{1}{2}\times2\times5^2 = 25,\text{J}).\n\nA4. (\text{PE}{\text{spring}} = \frac{1}{2} k x^2 = 0.5\times200\times0.1^2 = 1,\text{J}).\n\nA5. Work: (m g h = 100\times9.8\times20 = 19600,\text{J}).\n\n Power: (P = \frac{19600}{10} = 1960,\text{W} \approx 1.96,\text{kW}).\n\nA6. Initial speed 72 km/h = 20 m/s.\n\n (\text{KE}i = \frac{1}{2}\times1000\times20^2 = 2.0\times10^5,\text{J}).\n\n Final KE = 0, so brakes do (-2.0\times10^5,\text{J}) (magnitude 200 kJ).\n\nA7. \n(\displaystyle W = \int{0}^{2} 3x^2,dx = 3\left[\frac{x^3}{3}\right]{0}^{2} = (2^3) = 8,\text{J}).\n\nA8. (W = P t = 5000\times(3\times60) = 9.0\times10^5,\text{J}).\n\nA9. By work–energy theorem, if net work = 0 then change in KE = 0, so KE stays constant.\n\nA10. Height drop (h = L(1 - \cos\theta) = 2(1-\cos60^{\circ}) = 1,\text{m}).\n\n (m g h = \frac{1}{2} m v^2\Rightarrow v = \sqrt{2 g h} = \sqrt{2\times9.8\times1} \approx 4.4,\text{m/s}).\n\nFinal Answers (short): 1) 20 J, 2) 20.4 m, 3) 25 J, 4) 1 J, 5) 1960 W, 6) 2×10^5 J (opposite), 7) 8 J, 8) 9×10^5 J, 9) KE unchanged, 10) 4.4 m/s.