Mechanical Power

Q 1 :

A body is moving unidirectionally under the influence of a constant power source. Its displacement in time $t$ is proportional to [2024]

$t$
$t^{2}$
$t^{2 / 3}$
$t^{3 / 2}$

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(4)

$\frac{d W}{d t} = P \Rightarrow dW = P d t$

$F \cdot d s = P d t$

$m d v \frac{d s}{d t} = P d t$

$\int_{0}^{v} v d v = \frac{P}{m} \int_{0}^{t} d t \Rightarrow \frac{v^{2}}{2} = \frac{P t}{m}$

Hence, $v \propto \sqrt{t} \Rightarrow \frac{d s}{d t} = C \sqrt{t},$ $C$ is a constant

$\int_{0}^{s} d s = C \int_{0}^{t} t^{1 / 2} d t$

$s = C \frac{2}{3} t^{3 / 2} \Rightarrow s \propto t^{3 / 2}$

$∴$ displacement of proportional to ${(t)}^{3 / 2}$

Q 2 :

A body of mass 2 kg begins to move under the action of a time dependent force given by $\vec{F} = (6 t \hat{i} + 6 t^{2} \hat{j}) N .$ The power developed by the force at the time $t$ is given by [2024]

$(6 t^{4} + 9 t^{5}) W$
$(3 t^{5} + 6 t^{5}) W$
$(9 t^{5} + 6 t^{3}) W$
$(9 t^{3} + 6 t^{5}) W$

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(4)

$\vec{F} = (6 t \hat{i} + 6 t^{2} \hat{j}) N$

$\vec{F} = m \vec{a} = (6 t \hat{i} + 6 t^{2} \hat{j})$

$\vec{a} = \frac{\vec{F}}{m} = (3 t \hat{i} + 3 t^{2} \hat{j})$

$\vec{v} = \int_{0}^{t} \vec{a} d t = \frac{3 t^{2}}{2} \hat{i} + t^{3} \hat{j}$

$P = \vec{F} \cdot \vec{v} = (9 t^{3} + 6 t^{5}) W$

Q 3 :

A body of mass 4 kg is placed on a plane at a point P having coordinate (3, 4) m. Under the action of force $\vec{F} = (2 \hat{i} + 3 \hat{j}) N$ , it moves to a new point Q having coordinates (6, 10) m in 4 sec. The average power and instantaneous power at the end of 4 sec are in the ratio of : [2025]

13 : 6
6 : 13
1 : 2
4 : 13

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(2)

$\vec{F} = (2 \hat{i} + 3 \hat{j}) N$ ; displacement $\vec{S} = (3 \hat{i} + 6 \hat{j}) m$

Acceleration, $\vec{a} = \frac{\vec{F}}{m} = (\frac{2 \hat{i} + 3 \hat{j}}{4}) m / s^{2}$

$\vec{S} = \vec{u} t + \frac{1}{2} \vec{a} t^{2}$

$\Rightarrow 3 \hat{i} + 6 \hat{j} = 4 \vec{u} + \frac{1}{2} (\frac{2 \hat{i} + 3 \hat{j}}{4}) \cdot 16 = 4 \vec{u} + 4 \hat{i} + 6 \hat{j}$

or $\vec{u} = \frac{1}{4} (3 \hat{i} + 6 \hat{j} - 4 \hat{i} - 6 \hat{j}) \Rightarrow \vec{u} = - \frac{1}{4} \hat{i}$

Average Power $< P > = \vec{F} \cdot \frac{\vec{S}}{t}$

$= \frac{(2 \hat{i} + 3 \hat{j}) \cdot (3 \hat{i} + 6 \hat{j})}{4} watt = 6 watt$

$\vec{v}$ $at t = 4 \sec = (\frac{1}{2} \hat{i} + \frac{3}{4} \hat{j}) \times 4 = (2 \hat{i} + 3 \hat{j})$

$P_{ins} = \vec{F} \cdot \vec{v} = (2 \hat{i} + 3 \hat{j}) (2 \hat{i} + 3 \hat{j}) = 13 W$

$\Rightarrow \frac{< P >}{P_{ins}} = \frac{6}{13}$

Q 4 :

A sand dropper drops sand of mass m(t) on a conveyer belt at a rate proportional to the square root of speed (v) of the belt, i.e., $\frac{d m}{d t} \propto \sqrt{v}$ . If P is the power delivered to run the belt at constant speed then which of the following relationship is true? [2025]

$P^{2} \propto v^{3}$
$P \propto \sqrt{v}$
$P \propto v$
$P^{2} \propto v^{5}$

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(4)

Power = $\vec{F} \cdot \vec{v} and F = \frac{d p}{d t} = Rate of change of linear momentum$

$F = v \cdot \frac{d m}{d t} = K v^{3 / 2}, K is constant$

Power $(P) = {(K v)}^{3 / 2} \cdot (v) = K v^{5 / 2}$

So, $P^{2} \propto v^{5}$

Q 5 :

An object of mass 1000 g experiences a time dependent force $\vec{F} = (2 t \hat{i} + 3 t^{2} \hat{j}) N$ . The power generated by the force at time t is: [2025]

$(2 t^{3} + 3 t^{3}) W$
$(2 t^{2} + 18 t^{3}) W$
$(3 t^{3} + 5 t^{5}) W$
$(2 t^{3} + 3 t^{5}) W$

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(4)

m = 1000 gram = 1 kg

$\vec{F} = (2 t \hat{i} + 3 t^{2} \hat{j})$

$\frac{d v}{d t} = \frac{F}{m} = (2 t \hat{i} + 3 t^{2} \hat{j})$

$\Rightarrow \int_{0}^{v} d v = \int_{0}^{t} (2 t \hat{i} + 3 t^{2} \hat{j}) d t$

$\Rightarrow \vec{v} = t^{2} \hat{i} + t^{3} \hat{j}$

Power, $P = \vec{F} \cdot \vec{v}$

$P = (2 t \hat{i} + 3 t^{2} \hat{j}) \cdot (t^{2} \hat{i} + t^{3} \hat{j})$

$P = (2 t^{3} + 3 t^{5}) W$

Q 6 :

The ratio of powers of two motors is $\frac{3 \sqrt{x}}{\sqrt{x} + 1}$ , that are capable of raising 300 kg water in 5 minutes and 50 kg water in 2 minutes respectively from a well of 100 m deep. The value of $x$ will be [2023]

2
4
2.4
16

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(4)

$Average Power = \frac{total work done}{total time} = \frac{m g h}{t}$

$\frac{P_{1}}{P_{2}} = \frac{\frac{m_{1} g h}{t_{1}}}{\frac{m_{2} g h}{t_{2}}} = \frac{m_{1}}{t_{1}} \frac{t_{2}}{m_{2}}$

$\frac{P_{1}}{P_{2}} = \frac{300 \times 2}{5 \times 50} = \frac{12}{5} = \frac{3 \sqrt{x}}{\sqrt{x} + 1}$

$\Rightarrow x = 16$

Q 7 :

A body of mass $1 kg$ begins to move under the action of a time dependent force $\vec{F} = (t \hat{i} + 3 t^{2} \hat{j}) N.$ where $\hat{i}$ and $\hat{j}$ are the unit vectors along $x$ and $y$ axis. The power developed by the above force, at the time $t = 2 s$ will be _______ W. [2023]

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(100)

$\vec{F} = t \hat{i} + 3 t^{2} \hat{j}$

$\frac{m d \vec{v}}{d t} = t \hat{i} + 3 t^{2} \hat{j}$

$m = 1 kg$ , $\int_{0}^{v} d v = \int_{0}^{t} t d t \hat{i} + \int_{0}^{t} 3 t^{2} d t \hat{j}$

$\vec{v} = \frac{t^{2}}{2} \hat{i} + t^{3} \hat{j}$

$Power = \vec{F} \cdot \vec{v} = \frac{t^{3}}{2} + 3 t^{5}$

$At t = 2, Power = \frac{8}{2} + 3 \times 32 = 100$

Q 8 :

A body of mass 2 kg is initially at rest. It starts moving unidirectionally under the influence of a source of constant power P. Its displacement in 4 sec is $\frac{1}{3} α^{2} \sqrt{P} m .$ The value of $α$ will be ________ . [2023]

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(4)

$\frac{1}{2} m v^{2} = P t$

$v = \sqrt{\frac{2 P t}{m}}$

$\frac{d x}{d t} = \sqrt{\frac{2 P t}{m}}$

$x = \sqrt{\frac{2 P}{m}} \frac{2}{3} {[t^{3 / 2}]}_{0}^{4}$

$x = \frac{16 \sqrt{P}}{3}$ $= \frac{1}{3} \times 16 \sqrt{P}$

$\Rightarrow α = 4$

Q 9 :

If the maximum load carried by an elevator is 1400 kg (600 kg – passenger + 800 kg – elevator), which is moving up with a uniform speed of $3 {ms}^{- 1}$ and the frictional force acting on it is 2000 N, then the maximum power used by the motor is _______ kW $(g = 10 {m/s}^{2})$ [2023]

0%

(48)

$P_{\max} = F_{\max} \times v$

$F_{\max} = 1400 g + friction$

$= 14000 + 2000 = 16000$

$P_{\max} = 16000 \times 3 = 48000 W = 48 kW$

Q 10 :

A block of mass 5 kg starting from rest pulled up on a smooth inclined plane making an angle of $30 °$ with the horizontal with an effective acceleration of $1 {ms}^{- 2}$ . The power delivered by the pulling force at $t = 10 s$ from the start is ________ W. [Use $g = 10 {ms}^{- 2}$ ] (calculate the nearest integer value) [2023]

0%

(300)

$F - 5 \sin 30 ° = 5 a$

$\Rightarrow F = 5 + 25 = 30 N$

$v_{10} = u + a t$

$\Rightarrow v_{10} = 0 + 1 (10) = 10 m/s$

$P_{10} = F v = 300 W$

Topic Question Set

Q 1 :

A body is moving unidirectionally under the influence of a constant power source. Its displacement in time $t$ is proportional to [2024]

Q 2 :

A body of mass 2 kg begins to move under the action of a time dependent force given by $\vec{F} = (6 t \hat{i} + 6 t^{2} \hat{j}) N .$ The power developed by the force at the time $t$ is given by [2024]

Q 4 :

A sand dropper drops sand of mass m(t) on a conveyer belt at a rate proportional to the square root of speed (v) of the belt, i.e., $\frac{d m}{d t} \propto \sqrt{v}$ . If P is the power delivered to run the belt at constant speed then which of the following relationship is true? [2025]

Q 5 :

An object of mass 1000 g experiences a time dependent force $\vec{F} = (2 t \hat{i} + 3 t^{2} \hat{j}) N$ . The power generated by the force at time t is: [2025]

Q 6 :

The ratio of powers of two motors is $\frac{3 \sqrt{x}}{\sqrt{x} + 1}$ , that are capable of raising 300 kg water in 5 minutes and 50 kg water in 2 minutes respectively from a well of 100 m deep. The value of $x$ will be [2023]

Q 7 :

A body of mass $1 kg$ begins to move under the action of a time dependent force $\vec{F} = (t \hat{i} + 3 t^{2} \hat{j}) N.$ where $\hat{i}$ and $\hat{j}$ are the unit vectors along $x$ and $y$ axis. The power developed by the above force, at the time $t = 2 s$ will be _______ W. [2023]

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Q 8 :

A body of mass 2 kg is initially at rest. It starts moving unidirectionally under the influence of a source of constant power P. Its displacement in 4 sec is $\frac{1}{3} α^{2} \sqrt{P} m .$ The value of $α$ will be ________ . [2023]

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Q 9 :

If the maximum load carried by an elevator is 1400 kg (600 kg – passenger + 800 kg – elevator), which is moving up with a uniform speed of $3 {ms}^{- 1}$ and the frictional force acting on it is 2000 N, then the maximum power used by the motor is _______ kW $(g = 10 {m/s}^{2})$ [2023]

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