Work Energy & Power

Q: A body of mass m accelerates uniformly from rest to velocity v₀ in time t₀ , find the instantaneous power delivered to body when velocity is v₀/2.

Sol: Acceleration $\large a = \frac{v_0}{t_0} $

Force , $\large F = m \frac{v_0}{t_0} $

Instantaneous power , $\large P = F. \frac{v_0}{2} = (m \frac{v_0}{t_0})\frac{v_0}{2} $

$\large P = \frac{m v_0^2}{2 t_0} $

Q:  A hose pipe has a diameter of 2.5 cm and its required to direct a jet of water to a height of at least 40 m. Find the minimum power of the pump needed for this hose.

Sol: Volume of water ejected per sec is

$\large A v = \pi \times (\frac{d}{2})^2 \times \sqrt{2 g h}$ ; ( v = √(2gh))

Mass ejected per sec is $\large m = \frac{\pi d^2}{4}\times \sqrt{2 g h}\times \rho Kg/s $

Kinetic energy of water leaving hose/sec

$\large = \frac{1}{2}m v^2 = \frac{\pi d^2}{8}\times (2gh)^{3/2}\times \rho $

$\large = \frac{3.14 \times (2.5 \times 10^{-2})^2}{8}\times (2 \times 9.8 \times 40)^{3/2}\times 1000 $

= 21.5 KJ

Q: Find the power of an engine which can draw a train of 400 metric ton up the inclined plane of 1 in 98 at the rate 10 m/s. The resistance due to friction acting on the train is 10 N per ton.

Sol: Given sin θ = 1/98 ; m = 400 × 10³ kg

Frictional force f = 10 × 400 = 4000 N ;

Velocity v = 10 m/s

∴ power P = (mg sinθ + f) v

∴ P = [(400 × 10³ × 9.8 × 1/98)+ 4000] × 10

= 440000 W = 440 KW

Q: A machine delivers power to a body which is directly proportional to velocity of the body. If the body starts with a velocity which is almost negligible, find the distance covered by the body in attaining a velocity v .

Sol: $\large P \propto v $

$\large P = C v $ ; where C = constant

$\large m v \frac{dv}{dx} v = C v $

$\large m v \frac{dv}{dx} = C $

$\large \int_{0}^{v} v dv = \frac{C}{m} \int_{0}^{x} dx $

$\large \frac{v^2}{2} = \frac{C}{m} x $

$\large x = \frac{1}{2} \frac{m v^2}{C}$

Q: The potential energy of 1 kg particle free to move along X – axis is given by $U(x)= (\frac{x^4}{4} – \frac{x^2}{2}) J$ . The total mechanical energy of the particle is 2 J. Find the maximum speed of the particle.

Sol: For maximum value of U,

$\large \frac{dU}{dx} = 0$

$\large \frac{4 x^3}{4} -\frac{2 x}{2} = 0$

or , x = 0 , x = ±1

At x = 0 , $\large \frac{d^2U}{dx^2} = -1 $

and at x = ±1 , $\large \frac{d^2U}{dx^2} = 2 $

Hence U is minimum at x = ±1 with value U_min = -1/4 J

$\large E = K_{max} + U_{min}$

$\large 2 = K_{max} – \frac{1}{4} $

$\large K_{max} = \frac{9}{4} $

$\large \frac{1}{2} m v^2 = \frac{9}{4} $

$\large v_{max} = \frac{3}{\sqrt{2}} m/s$

Q: A particle is released from height H. At certain height from the ground its kinetic energy is twice its gravitational potential energy. Find the height and speed of particle at that height

Sol: K.E = 2PE but KE = TE – PE

mg(H – h) = 2mgh

mg H = 3 mgh

⇒ h = H/3

Also K.E = 2P.E

$\large \frac{1}{2}m v^2 = 2 mg h $

$\large \frac{1}{2}m v^2 = 2 mg (\frac{H}{3}) $

$\large v = 2 \sqrt{\frac{g H}{3}}$

Q: A vehicle of mass 15 quintal climbs up a hill 200 m high. It then moves on a level road with a speed of 30 m/s. Calculate the potential energy gained by it and its total mechanical energy while running on the top of the hill.

Sol: m = 15 quintal = 1500 kg , g = 9.8ms^-2, h = 200 m

P.E. gained, U = mgh = 1500 × 9.8 × 200 = 2.94 × 10⁶ J

K.E. = (1/2 ) m v² = (1/2) × 1500 × (30)² = 0.675 × 10⁶ J

Total mechanical energy

E = K + U = (0.675 + 2.94) × 10⁶ = 3.615 × 10⁶ J