Question 41:
easyA body projected vertically from the earth reaches a height equal to earth’s radius before returning to the earth. The power exerted by the gravitational force is greatest
(2011 Pre)
Power is $P = \vec{F} \cdot \vec{v} = Fv$. The gravitational force is downward and speed is maximum at the instant just before the body hits the earth, making the power exerted by gravity greatest at that moment.
Question 42:
moderateA particle of mass $M$ starting from rest undergoes uniform acceleration. If the speed acquired in time $T$ is $V$, the power delivered to the particle is:
(2010 Mains)
Power delivered is the rate of change of kinetic energy: $P = \frac{\Delta K}{T} = \frac{\frac{1}{2}MV^2 - 0}{T} = \frac{1}{2}\frac{MV^2}{T}$.
Question 43:
moderateAn engine pumps water through a hose pipe. Water passes through the pipe and leaves it with a velocity of $2\text{ m/s}$. The mass per length of water in the pipe is $100\text{ kg/m}$. What is the power of the engine?
(2010 Pre)
Mass flow rate is $\frac{dm}{dt} = \lambda v = 100 \times 2 = 200\text{ kg/s}$. Power of the engine equals the kinetic energy given to water per second: $P = \frac{1}{2}\left(\frac{dm}{dt}\right)v^2 = \frac{1}{2}(200)(2)^2 = 400\text{ W}$.
Question 44:
moderateA block of mass $10\text{ kg}$ moving in $x$ direction with a constant speed of $10\text{ m s}^{-1}$, is subjected to a retarding force $F = -0.1x\text{ J/m}$ during its travel from $x = 20\text{ m}$ to $30\text{ m}$. Its final K.E. will be:
(2015)
Use work-energy theorem: $K_f - K_i = W$. Initial kinetic energy is $K_i = \frac{1}{2}mv^2 = 500\text{ J}$. Work done by the force is $W = \int_{20}^{30} (-0.1x) dx = -25\text{ J}$. Thus, final kinetic energy is $K_f = 500 - 25 = 475\text{ J}$.
Question 45:
moderateA body of mass $1\text{ kg}$ is thrown upwards with a velocity $20\text{ m/s}$. It momentarily comes to rest after attaining a height of $18\text{ m}$. How much energy is lost due to air friction? ($g = 10\text{ m/s}^2$)
(2009)
Apply conservation of energy with non-conservative forces: initial energy $E_i = \frac{1}{2}mv^2 = 200\text{ J}$. Final potential energy at $18\text{ m}$ is $U_f = mgh = 180\text{ J}$. Energy lost against air friction is $\Delta E = E_i - U_f = 200 - 180 = 20\text{ J}$.
Question 46:
easyIf $\vec{F} = (60\hat{i} + 15\hat{j} – 3\hat{k}) \text{ N}$ and $\vec{V} = (2\hat{i} – 4\hat{j} + 5\hat{k}) \text{ m/s}$ then instantaneous power is:
(2000)
Instantaneous power is given by $P = \vec{F} \cdot \vec{V} = (60)(2) + (15)(-4) + (-3)(5) = 120 - 60 - 15 = 45 \text{ W}$.
Question 47:
moderateHow much water a pump of $2 \text{ kW}$ can raise in one minute to a height of $10 \text{ m}$? (Take $g = 10 \text{ m/s}^2$)
(1990)
Using $P = \frac{mgh}{t}$, mass $m = \frac{P \cdot t}{gh} = \frac{2000 \times 60}{10 \times 10} = 1200 \text{ kg}$, which equals $1200 \text{ litres}$.
Question 48:
moderateAn engine pumps water continuously through a hose. Water leaves the hose with a velocity $v$ and $m$ is the mass per unit length of the water jet. What is the rate at which kinetic energy is imparted to water?
(2009)
Mass flowing per second is $\frac{dm}{dt} = mv$. The rate of kinetic energy is $\frac{1}{2} \left(\frac{dm}{dt}\right) v^2 = \frac{1}{2} mv^3$.
Question 49:
easy250 N force is required to raise 75 kg mass from a pulley. If rope is pulled 12 m then the load is lifted to 3 m, the efficiency of pulley system will be:
(2001)
Efficiency $\eta = \frac{\text{Work output}}{\text{Work input}} = \frac{mgh}{F \cdot d} = \frac{75 \times 10 \times 3}{250 \times 12} = \frac{3}{4} = 75\%$.