Questions: Work, Energy and Power

Q1. Define work and state when it is zero, positive and negative.

Answer: Work is the scalar product of force and displacement, so it equals F d cos theta for a constant force. It is zero when the force is perpendicular to the displacement, as for the centripetal force or when a person carries a load horizontally at constant speed. It is positive when the force has a component along the displacement, and negative when that component is opposite, as for friction and for gravity on a rising body. The unit is the joule.

Hinglish: Work F d cos theta hai. Zero tab hota hai jab force displacement ke perpendicular ho, jaise circular motion mein centripetal force. Positive tab jab force ka component displacement ke along ho, aur negative tab jab opposite ho, jaise friction ya upar jaate body par gravity. Unit joule hai.

Q2. State and prove the work-energy theorem for a constant force.

Answer: The theorem says the net work on a body equals its change in kinetic energy. For a constant force along the displacement, v squared equals u squared plus 2 a s, and multiplying by mass over two gives half m v squared minus half m u squared equal to m a s, which is the work. The proof for a varying force is the same idea with the integral of force over displacement, using the second law at each point. The theorem connects forces and speeds without using time.

Hinglish: Theorem kehta hai net work change in kinetic energy ke barabar hai. Constant force ke liye v^2 = u^2 + 2as, isko m/2 se multiply karne par work ka expression milta hai. Variable force ke liye force ka integral lena padta hai. Is theorem se time nikaale bina force aur speed ka rishta mil jaata hai.

Q3. A force of 20 N acts at 60 degrees to the displacement of 5 m. Find the work done.

Answer: The work is F d cos theta, so 20 times 5 times the cosine of 60 degrees. The cosine of 60 is 0.5, so the work is 50 J. Only the component along the displacement, 10 N, does work; the perpendicular component of about 17.3 N does none. If the same force were at 90 degrees, the work would be zero however large the force.

Hinglish: Work F d cos theta = 20 x 5 x cos 60 = 50 J. Sirf displacement ke along component 10 N kaam karta hai; perpendicular component ka work zero. Agar angle 90 degree hota to force kitni bhi ho, work zero hota.

Q4. What is kinetic energy? Show its relation with momentum.

Answer: Kinetic energy measures motion energy, and its value is half the mass multiplied by the speed squared. Writing momentum as p equal to m v turns this into p squared over twice the mass. This form shows that for a fixed momentum the kinetic energy is smaller for the heavier body, and for a fixed kinetic energy the heavier body has more momentum. The relation is useful in collision problems where momentum is conserved but kinetic energy is not.

Hinglish: Kinetic energy 1/2 m v^2 hoti hai. Momentum p = mv se likho to K = p^2/(2m). Fixed momentum par bhaari body ki kinetic energy kam hoti hai, aur fixed kinetic energy par bhaari body ka momentum zyada. Collision problems mein yahi relation kaam aata hai.

Q5. Distinguish between conservative and non conservative forces with examples.

Answer: A conservative force does work that depends only on the initial and final positions, and the work around any closed path is zero. Gravity, the spring force and the electrostatic force are conservative. A non conservative force does work that depends on the path, and the work around a closed loop is not zero; friction and air resistance are the standard examples. Only conservative forces have a potential energy associated with them, which is why friction problems need an explicit energy loss term.

Hinglish: Conservative force ka work sirf initial aur final position par depend karta hai, closed path par zero hota hai, jaise gravity aur spring. Non conservative force ka work path par depend karta hai, jaise friction; closed loop mein bhi energy kharch hoti hai. Sirf conservative forces ke liye potential energy hoti hai.

Q6. A ball is dropped from 20 m. Find its speed when it reaches the ground, using energy conservation.

Answer: The initial energy is all potential, m g h, and the final energy is all kinetic, half m v squared. Setting them equal cancels the mass and gives v squared equal to 2 g h, which is 400 with g equal to 10, so v is 20 m per second. The same result follows from the kinematic equation, but the energy route needs no time and no sign convention beyond fixing the zero of potential energy at the ground.

Hinglish: Shuru mein energy sirf potential mgh, neeche pahunchne par sirf kinetic 1/2 m v^2. Mass cancel, v^2 = 2gh = 400, matlab v = 20 m/s. Kinematics se bhi wahi answer aata hai, par energy method mein time ki zaroorat nahi.

Q7. Define power, give its unit and relate it to force and velocity.

Answer: Power is the rate of doing work, the work divided by the time for an average, and its SI unit is the watt, one joule per second. For a force acting on a moving body, the instantaneous power is the scalar product of force and velocity. A car engine of fixed power therefore produces a larger force at low speed and a smaller force at high speed, because the product of the two is fixed. A kilowatt hour is a unit of energy, not power, equal to 3.6 million joule.

Hinglish: Power work ki rate hai, P = W/t, unit watt. Moving body ke liye P = F v hota hai. Fixed power wala engine low speed par zyada force deta hai aur high speed par kam, kyunki dono ka product fixed hai. Kilowatt hour energy ki unit hai, power ki nahi, aur woh 3.6 x 10^6 J hota hai.

Q8. A spring of constant 200 N m^-1 is compressed by 10 cm. Find the energy stored.

Answer: The energy stored in a spring is half the spring constant times the square of the deformation. Here k is 200 and x is 0.1 m, so the energy is half of 200 times 0.01, which is 1 J. The same energy is released when the spring returns to its natural length, whatever the direction of the original deformation, because the energy depends on x squared. A block launched by this spring of mass 0.5 kg would leave with a speed given by half m v squared equal to 1, so v is 2 m per second.

Hinglish: Spring ki energy 1/2 k x^2 hoti hai. Yahan k = 200, x = 0.1 m, isliye energy 1/2 x 200 x 0.01 = 1 J. Compression aur extension dono ke liye energy same hoti hai kyunki x squared hai. 0.5 kg ke block ko yeh energy 2 m/s ki speed de degi.

Q9. Distinguish between elastic and inelastic collisions.

Answer: In an elastic collision both momentum and kinetic energy are conserved, and the relative speed of approach equals the relative speed of separation. In an inelastic collision momentum is conserved but kinetic energy is not, part of it appearing as heat, sound and deformation. A perfectly inelastic collision is the extreme case where the bodies stick together and move with a common velocity. Elastic collisions occur between hard spheres in ideal conditions; most real collisions are partly inelastic.

Hinglish: Elastic collision mein momentum aur kinetic energy dono conserved, aur approach speed separation speed ke barabar. Inelastic mein momentum conserved par kinetic energy nahi, kuch energy heat aur deformation mein chali jaati hai. Perfectly inelastic mein bodies chipak kar common velocity se chalti hain. Real collisions zyadatar partly inelastic hote hain.

Q10. A 2 kg body moving at 6 m s^-1 collides with a 4 kg body at rest and they stick together. Find the common velocity and the loss of kinetic energy.

Answer: Momentum is conserved, so the total momentum 12 kg m per second is shared by the combined mass of 6 kg, giving a common velocity of 2 m per second. The initial kinetic energy is half of 2 times 36, that is 36 J, and the final kinetic energy is half of 6 times 4, that is 12 J. The loss is 24 J, which appears as heat and deformation. The final kinetic energy is positive and smaller than the initial, as it must be for an inelastic collision.

Hinglish: Momentum conserve: 2 x 6 = 12 kg m/s, combined mass 6 kg, velocity 12/6 = 2 m/s. Initial KE 1/2 x 2 x 36 = 36 J, final KE 1/2 x 6 x 4 = 12 J, loss 24 J. Yeh loss heat aur deformation mein chala jaata hai, aur final KE initial se kam hai, jo inelastic collision mein zaroori hai.

Quiz below: five questions on work, energy conservation, power and collisions. Attempt them before rereading the formula sheet, and check each wrong answer against the matching page.