NEET Work Energy and Power MCQs & PYQs

Practice NEET Work Energy and Power Questions

Question 31:

moderate

A body of mass $3\text{ kg}$ is under a constant force which causes a displacement $s$ in meters in it, given by the relation $s = \frac{1}{3} t^2$, where $t$ is in $s$. Work done by the force in $2\text{ s}$ is:

(2006)

Find acceleration by differentiating displacement twice: $v = \frac{2}{3}t$ and $a = \frac{2}{3}\text{ m/s}^2$. Force is $F = ma = 2\text{ N}$. At $t = 2\text{ s}$, displacement is $s = \frac{4}{3}\text{ m}$. Work done $W = F \times s = \frac{8}{3}\text{ J}$.

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Question 32:

easy

If $x = 3 – 4t^2 + t^3$, then work done in first $4\text{ s}$ will be (Mass of the particle is $3\text{ gram}$):

(1998)

Use work-energy theorem $W = \Delta K = \frac{1}{2}m(v_f^2 - v_i^2)$. Velocity is $v = -8t + 3t^2$. At $t = 0$, $v_i = 0$ and at $t = 4\text{ s}$, $v_f = 16\text{ m/s}$. Substituting values with $m = 3\text{ g} = 0.003\text{ kg}$ gives $W = 384\text{ mJ}$.

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Question 33:

moderate

A bullet of mass $10\text{ g}$ leaves a rifle at an initial velocity of $100\text{ m/s}$ and strikes the earth at the same level with a velocity of $500\text{ m/s}$. The work done in joule overcoming the resistance of air will be:

(1989)

By work-energy theorem, the work done by air resistance plus gravity equals change in kinetic energy. Since it returns to the same level, net work by gravity is zero. Thus, work done by air resistance is $W = \Delta K = \frac{1}{2}m(v^2 - u^2) = 3750\text{ J}$.

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Question 34:

moderate

An electric lift with a maximum load of $2000\text{ kg}$ (lift + passengers) is moving up with a constant speed of $1.5\text{ ms}^{-1}$. The frictional force opposing the motion is $3000\text{ N}$. The minimum power delivered by the motor to the lift in watts is : ($g = 10\text{ ms}^{-2}$)

(2022)

Total upward force required equals the sum of weight and frictional force: $F = mg + f = (2000 \times 10) + 3000 = 23000\text{ N}$. Minimum power $P = F \cdot v = 23000 \times 1.5 = 34500\text{ W}$.

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Question 35:

easy

The energy that will be ideally radiated by a $100\text{ kW}$ transmitter in $1\text{ hour}$ is:

(2022)

Energy is given by $E = P \times t$. Here, power $P = 100\text{ kW} = 10^5\text{ W}$ and time $t = 1\text{ hour} = 3600\text{ s}$. Thus, $E = 10^5 \times 3600 = 3.6 \times 10^8\text{ J} = 36 \times 10^7\text{ J}$.

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Question 36:

moderate

Water falls from a height of $60\text{ m}$ at the rate of $15\text{ kg/s}$ to operate a turbine. The losses due to frictional force are $10\%$ of the input energy. How much power is generated by the turbine? ($g = 10\text{ m/s}^2$)

(2021, 2008)

Input power is $P_{\text{in}} = \left(\frac{dm}{dt}\right)gh = 15 \times 10 \times 60 = 9000\text{ W} = 9\text{ kW}$. With $10\%$ frictional losses, efficiency is $90\%$. Power generated = $90\% \text{ of } 9\text{ kW} = 8.1\text{ kW}$.

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Question 37:

moderate

A body of mass $1\text{ kg}$ begins to move under the action of a time dependent force $\vec{F} = (2t\hat{i} + 3t^2\hat{j})\text{ N}$, where $\hat{i}$ and $\hat{j}$ are unit vectors along $x$ and $y$ axis. What power will be developed by the force at the time $t$?

(2016 – I)

Velocity is found by integrating acceleration: $\vec{v} = \int \frac{\vec{F}}{m} dt = (t^2\hat{i} + t^3\hat{j})$. Power is given by the dot product $P = \vec{F} \cdot \vec{v} = (2t)(t^2) + (3t^2)(t^3) = 2t^3 + 3t^5\text{ W}$.

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Question 38:

moderate

A particle of mass $m$ is driven by a machine that delivers a constant power $k$ watts. If the particle starts from rest the force on the particle at time $t$ is:

(2015)

Using constant power $P = Fv$ and $P = mv\frac{dv}{dt}$, integrating gives velocity $v = \sqrt{\frac{2kt}{m}}$. Since force $F = \frac{P}{v}$, substituting velocity yields $F = \sqrt{\frac{mk}{2}} t^{-1/2}$.

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Question 39:

moderate

The heart of a man pumps $5\text{ litres}$ of blood through the arteries per minute at a pressure of $150\text{ mm}$ of mercury. If the density of mercury be $13.6 \times 10^3\text{ kg/m}^3$ and $g = 10\text{ m/s}^2$, then the power of heart in watt is:

(2015 Re)

Power is given by $P = p \left(\frac{dV}{dt}\right)$. Pressure $p = h\rho g = 0.15 \times 13.6 \times 10^3 \times 10 = 2.04 \times 10^4\text{ Pa}$. Volume flow rate $\frac{dV}{dt} = \frac{5 \times 10^{-3}}{60}\text{ m}^3\text{/s}$. Multiplying these gives $1.70\text{ W}$.

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Question 40:

moderate

A car of mass $m$ starts from rest and accelerates so that the instantaneous power delivered to the car has a constant magnitude $P_0$. The instantaneous velocity of this car is proportional to:

(2012 Mains)

From constant power $P_0 = mv\frac{dv}{dt}$, integrate both sides to get $v^2 = \frac{2P_0 t}{m}$. Thus, velocity $v \propto t^{1/2}$.

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