355518
A block of mass \(m\) is allowed to slide down a fixed smooth inclined plane of angle \(\theta\) and length \(l\). What is the magnitude of power developed by the gravitational force when the block reaches the bottom?
1 \(\sqrt{2 m^{2} l(g \sin \theta)^{3}}\)
2 \(\sqrt{(2 / 3) m^{3} l \mathrm{~g}^{2} \sin \theta}\)
3 \(\sqrt{(2 / 3) m^{2} l^{2} g \cos \theta}\)
4 \(\sqrt{(1 / 3) m^{3} l \mathrm{~g}^{2} \sin \theta}\)
Explanation:
The force that accelerates the block is \(m g \sin \theta \Rightarrow a=g \sin \theta\) Velocity of the block after travelling distance \(l\) is \(v^{2}=2 g \sin \theta l\) The instantaneous power is \(\begin{aligned}& P=F v=m g \sin \theta \sqrt{2 g \sin \theta l} \\& P=\sqrt{2 m^{2} g^{3} \sin ^{3} \theta l}\end{aligned}\)
PHXI06:WORK ENERGY AND POWER
355519
A particle moves with a velocity \(5\hat i - 3\hat j + 6\hat k\) \(m{\text{/}}s\) under the influence of a constant force \(F = 10\hat i + 10\hat j + 20\hat k\) \(N\). The instantaneous power applied to the particle is
355520
An engine exerts a force \(F = (20\widehat i - 3\widehat j + 5\widehat k)N\) and moves with velocity \(v = (6\hat i + 20\hat j - 3\hat k)m/s\). The power of the engine ( in watt) is
1 45
2 75
3 20
4 10
Explanation:
Here, Force, \(\vec F = (20\hat i - 3\hat j + 5\hat k)N\) Velocity, \(\vec v = (6\hat i + 20\hat j - 3\hat k)m/s\) \(P = \vec F \cdot \vec v = (20\hat i - 3\hat j + 5\hat k) \cdot (6\hat i + 20\hat j - 3\hat k)\) \(120 - 60 - 15 = 45W\)
PHXI06:WORK ENERGY AND POWER
355521
A body of mass \(2\,kg\) begins to move under the action of a time dependent force given by \(\vec{F}=\left(6 t \hat{i}+6 t^{2} \hat{j}\right) N\). The power developed by the force at this time \(t\) is given by
1 \(\left(3 t^{3}+6 t^{5}\right) W\)
2 \(\left(6 t^{4}+9 t^{5}\right) W\)
3 \(\left(9 t^{5}+6 t^{3}\right) W\)
4 \(\left(9 t^{3}+6 t^{5}\right) W\)
Explanation:
Here, \(\vec F = \left( {6t\hat i + 6{t^2}\hat j} \right)N;\,\,m = 2\;kg\) Acceleration of the body, \(\vec{a}=\dfrac{\vec{F}}{m}\) \(\vec a = \frac{{\left( {6t\hat i + 6{t^2}\hat j} \right)}}{2} = \left( {3t\hat i + 3{t^2}j} \right)m/{s^2}\) Velocity of the body at time \((t)\) is, \(\vec v = \int {\vec a} dt = \int {\left( {3t\hat i + 3{t^2}\hat j} \right)} \,dt\) \( = \left( {\frac{{3{t^2}}}{2}\hat i + \frac{{3{t^3}}}{3}\hat j} \right) = \left( {\frac{{3{t^2}}}{2}\hat i + {t^3}\hat j} \right)m/s\) Power developed by the force at time ' \(t\) ' \(P=\vec{F} \cdot \vec{v}=\left(6 t \hat{i}+6 t^{2} \hat{j}\right) \cdot\left(\dfrac{3 t^{2}}{2} \hat{i}+t^{3} \hat{j}\right)\) \(P=\left(9 t^{3}+6 t^{5}\right) W\)
JEE - 2024
PHXI06:WORK ENERGY AND POWER
355522
A car of mass \(m\) starts from rest and accelerates so that instantaneous power delivered to the car has a constant magnitude \(P_{0}\). The instantaneous velocity of this car is proportional to
1 \(\dfrac{t}{\sqrt{m}}\)
2 \(t^{2} P_{0}\)
3 \(t^{1 / 2}\)
4 \(t^{-1 / 2}\)
Explanation:
Constant power of a car \(P_{0}=F \cdot v=\) ma. \(v\) \(\begin{gathered}P_{0}=m \dfrac{d v}{d t} \cdot v \Rightarrow P_{0} d t=m v d v \Rightarrow P_{0} . t=\dfrac{m v^{2}}{2} \\v=\sqrt{\dfrac{2 P_{0} t}{m}} \Rightarrow v \alpha \sqrt{t} .\end{gathered}\)
355518
A block of mass \(m\) is allowed to slide down a fixed smooth inclined plane of angle \(\theta\) and length \(l\). What is the magnitude of power developed by the gravitational force when the block reaches the bottom?
1 \(\sqrt{2 m^{2} l(g \sin \theta)^{3}}\)
2 \(\sqrt{(2 / 3) m^{3} l \mathrm{~g}^{2} \sin \theta}\)
3 \(\sqrt{(2 / 3) m^{2} l^{2} g \cos \theta}\)
4 \(\sqrt{(1 / 3) m^{3} l \mathrm{~g}^{2} \sin \theta}\)
Explanation:
The force that accelerates the block is \(m g \sin \theta \Rightarrow a=g \sin \theta\) Velocity of the block after travelling distance \(l\) is \(v^{2}=2 g \sin \theta l\) The instantaneous power is \(\begin{aligned}& P=F v=m g \sin \theta \sqrt{2 g \sin \theta l} \\& P=\sqrt{2 m^{2} g^{3} \sin ^{3} \theta l}\end{aligned}\)
PHXI06:WORK ENERGY AND POWER
355519
A particle moves with a velocity \(5\hat i - 3\hat j + 6\hat k\) \(m{\text{/}}s\) under the influence of a constant force \(F = 10\hat i + 10\hat j + 20\hat k\) \(N\). The instantaneous power applied to the particle is
355520
An engine exerts a force \(F = (20\widehat i - 3\widehat j + 5\widehat k)N\) and moves with velocity \(v = (6\hat i + 20\hat j - 3\hat k)m/s\). The power of the engine ( in watt) is
1 45
2 75
3 20
4 10
Explanation:
Here, Force, \(\vec F = (20\hat i - 3\hat j + 5\hat k)N\) Velocity, \(\vec v = (6\hat i + 20\hat j - 3\hat k)m/s\) \(P = \vec F \cdot \vec v = (20\hat i - 3\hat j + 5\hat k) \cdot (6\hat i + 20\hat j - 3\hat k)\) \(120 - 60 - 15 = 45W\)
PHXI06:WORK ENERGY AND POWER
355521
A body of mass \(2\,kg\) begins to move under the action of a time dependent force given by \(\vec{F}=\left(6 t \hat{i}+6 t^{2} \hat{j}\right) N\). The power developed by the force at this time \(t\) is given by
1 \(\left(3 t^{3}+6 t^{5}\right) W\)
2 \(\left(6 t^{4}+9 t^{5}\right) W\)
3 \(\left(9 t^{5}+6 t^{3}\right) W\)
4 \(\left(9 t^{3}+6 t^{5}\right) W\)
Explanation:
Here, \(\vec F = \left( {6t\hat i + 6{t^2}\hat j} \right)N;\,\,m = 2\;kg\) Acceleration of the body, \(\vec{a}=\dfrac{\vec{F}}{m}\) \(\vec a = \frac{{\left( {6t\hat i + 6{t^2}\hat j} \right)}}{2} = \left( {3t\hat i + 3{t^2}j} \right)m/{s^2}\) Velocity of the body at time \((t)\) is, \(\vec v = \int {\vec a} dt = \int {\left( {3t\hat i + 3{t^2}\hat j} \right)} \,dt\) \( = \left( {\frac{{3{t^2}}}{2}\hat i + \frac{{3{t^3}}}{3}\hat j} \right) = \left( {\frac{{3{t^2}}}{2}\hat i + {t^3}\hat j} \right)m/s\) Power developed by the force at time ' \(t\) ' \(P=\vec{F} \cdot \vec{v}=\left(6 t \hat{i}+6 t^{2} \hat{j}\right) \cdot\left(\dfrac{3 t^{2}}{2} \hat{i}+t^{3} \hat{j}\right)\) \(P=\left(9 t^{3}+6 t^{5}\right) W\)
JEE - 2024
PHXI06:WORK ENERGY AND POWER
355522
A car of mass \(m\) starts from rest and accelerates so that instantaneous power delivered to the car has a constant magnitude \(P_{0}\). The instantaneous velocity of this car is proportional to
1 \(\dfrac{t}{\sqrt{m}}\)
2 \(t^{2} P_{0}\)
3 \(t^{1 / 2}\)
4 \(t^{-1 / 2}\)
Explanation:
Constant power of a car \(P_{0}=F \cdot v=\) ma. \(v\) \(\begin{gathered}P_{0}=m \dfrac{d v}{d t} \cdot v \Rightarrow P_{0} d t=m v d v \Rightarrow P_{0} . t=\dfrac{m v^{2}}{2} \\v=\sqrt{\dfrac{2 P_{0} t}{m}} \Rightarrow v \alpha \sqrt{t} .\end{gathered}\)
355518
A block of mass \(m\) is allowed to slide down a fixed smooth inclined plane of angle \(\theta\) and length \(l\). What is the magnitude of power developed by the gravitational force when the block reaches the bottom?
1 \(\sqrt{2 m^{2} l(g \sin \theta)^{3}}\)
2 \(\sqrt{(2 / 3) m^{3} l \mathrm{~g}^{2} \sin \theta}\)
3 \(\sqrt{(2 / 3) m^{2} l^{2} g \cos \theta}\)
4 \(\sqrt{(1 / 3) m^{3} l \mathrm{~g}^{2} \sin \theta}\)
Explanation:
The force that accelerates the block is \(m g \sin \theta \Rightarrow a=g \sin \theta\) Velocity of the block after travelling distance \(l\) is \(v^{2}=2 g \sin \theta l\) The instantaneous power is \(\begin{aligned}& P=F v=m g \sin \theta \sqrt{2 g \sin \theta l} \\& P=\sqrt{2 m^{2} g^{3} \sin ^{3} \theta l}\end{aligned}\)
PHXI06:WORK ENERGY AND POWER
355519
A particle moves with a velocity \(5\hat i - 3\hat j + 6\hat k\) \(m{\text{/}}s\) under the influence of a constant force \(F = 10\hat i + 10\hat j + 20\hat k\) \(N\). The instantaneous power applied to the particle is
355520
An engine exerts a force \(F = (20\widehat i - 3\widehat j + 5\widehat k)N\) and moves with velocity \(v = (6\hat i + 20\hat j - 3\hat k)m/s\). The power of the engine ( in watt) is
1 45
2 75
3 20
4 10
Explanation:
Here, Force, \(\vec F = (20\hat i - 3\hat j + 5\hat k)N\) Velocity, \(\vec v = (6\hat i + 20\hat j - 3\hat k)m/s\) \(P = \vec F \cdot \vec v = (20\hat i - 3\hat j + 5\hat k) \cdot (6\hat i + 20\hat j - 3\hat k)\) \(120 - 60 - 15 = 45W\)
PHXI06:WORK ENERGY AND POWER
355521
A body of mass \(2\,kg\) begins to move under the action of a time dependent force given by \(\vec{F}=\left(6 t \hat{i}+6 t^{2} \hat{j}\right) N\). The power developed by the force at this time \(t\) is given by
1 \(\left(3 t^{3}+6 t^{5}\right) W\)
2 \(\left(6 t^{4}+9 t^{5}\right) W\)
3 \(\left(9 t^{5}+6 t^{3}\right) W\)
4 \(\left(9 t^{3}+6 t^{5}\right) W\)
Explanation:
Here, \(\vec F = \left( {6t\hat i + 6{t^2}\hat j} \right)N;\,\,m = 2\;kg\) Acceleration of the body, \(\vec{a}=\dfrac{\vec{F}}{m}\) \(\vec a = \frac{{\left( {6t\hat i + 6{t^2}\hat j} \right)}}{2} = \left( {3t\hat i + 3{t^2}j} \right)m/{s^2}\) Velocity of the body at time \((t)\) is, \(\vec v = \int {\vec a} dt = \int {\left( {3t\hat i + 3{t^2}\hat j} \right)} \,dt\) \( = \left( {\frac{{3{t^2}}}{2}\hat i + \frac{{3{t^3}}}{3}\hat j} \right) = \left( {\frac{{3{t^2}}}{2}\hat i + {t^3}\hat j} \right)m/s\) Power developed by the force at time ' \(t\) ' \(P=\vec{F} \cdot \vec{v}=\left(6 t \hat{i}+6 t^{2} \hat{j}\right) \cdot\left(\dfrac{3 t^{2}}{2} \hat{i}+t^{3} \hat{j}\right)\) \(P=\left(9 t^{3}+6 t^{5}\right) W\)
JEE - 2024
PHXI06:WORK ENERGY AND POWER
355522
A car of mass \(m\) starts from rest and accelerates so that instantaneous power delivered to the car has a constant magnitude \(P_{0}\). The instantaneous velocity of this car is proportional to
1 \(\dfrac{t}{\sqrt{m}}\)
2 \(t^{2} P_{0}\)
3 \(t^{1 / 2}\)
4 \(t^{-1 / 2}\)
Explanation:
Constant power of a car \(P_{0}=F \cdot v=\) ma. \(v\) \(\begin{gathered}P_{0}=m \dfrac{d v}{d t} \cdot v \Rightarrow P_{0} d t=m v d v \Rightarrow P_{0} . t=\dfrac{m v^{2}}{2} \\v=\sqrt{\dfrac{2 P_{0} t}{m}} \Rightarrow v \alpha \sqrt{t} .\end{gathered}\)
355518
A block of mass \(m\) is allowed to slide down a fixed smooth inclined plane of angle \(\theta\) and length \(l\). What is the magnitude of power developed by the gravitational force when the block reaches the bottom?
1 \(\sqrt{2 m^{2} l(g \sin \theta)^{3}}\)
2 \(\sqrt{(2 / 3) m^{3} l \mathrm{~g}^{2} \sin \theta}\)
3 \(\sqrt{(2 / 3) m^{2} l^{2} g \cos \theta}\)
4 \(\sqrt{(1 / 3) m^{3} l \mathrm{~g}^{2} \sin \theta}\)
Explanation:
The force that accelerates the block is \(m g \sin \theta \Rightarrow a=g \sin \theta\) Velocity of the block after travelling distance \(l\) is \(v^{2}=2 g \sin \theta l\) The instantaneous power is \(\begin{aligned}& P=F v=m g \sin \theta \sqrt{2 g \sin \theta l} \\& P=\sqrt{2 m^{2} g^{3} \sin ^{3} \theta l}\end{aligned}\)
PHXI06:WORK ENERGY AND POWER
355519
A particle moves with a velocity \(5\hat i - 3\hat j + 6\hat k\) \(m{\text{/}}s\) under the influence of a constant force \(F = 10\hat i + 10\hat j + 20\hat k\) \(N\). The instantaneous power applied to the particle is
355520
An engine exerts a force \(F = (20\widehat i - 3\widehat j + 5\widehat k)N\) and moves with velocity \(v = (6\hat i + 20\hat j - 3\hat k)m/s\). The power of the engine ( in watt) is
1 45
2 75
3 20
4 10
Explanation:
Here, Force, \(\vec F = (20\hat i - 3\hat j + 5\hat k)N\) Velocity, \(\vec v = (6\hat i + 20\hat j - 3\hat k)m/s\) \(P = \vec F \cdot \vec v = (20\hat i - 3\hat j + 5\hat k) \cdot (6\hat i + 20\hat j - 3\hat k)\) \(120 - 60 - 15 = 45W\)
PHXI06:WORK ENERGY AND POWER
355521
A body of mass \(2\,kg\) begins to move under the action of a time dependent force given by \(\vec{F}=\left(6 t \hat{i}+6 t^{2} \hat{j}\right) N\). The power developed by the force at this time \(t\) is given by
1 \(\left(3 t^{3}+6 t^{5}\right) W\)
2 \(\left(6 t^{4}+9 t^{5}\right) W\)
3 \(\left(9 t^{5}+6 t^{3}\right) W\)
4 \(\left(9 t^{3}+6 t^{5}\right) W\)
Explanation:
Here, \(\vec F = \left( {6t\hat i + 6{t^2}\hat j} \right)N;\,\,m = 2\;kg\) Acceleration of the body, \(\vec{a}=\dfrac{\vec{F}}{m}\) \(\vec a = \frac{{\left( {6t\hat i + 6{t^2}\hat j} \right)}}{2} = \left( {3t\hat i + 3{t^2}j} \right)m/{s^2}\) Velocity of the body at time \((t)\) is, \(\vec v = \int {\vec a} dt = \int {\left( {3t\hat i + 3{t^2}\hat j} \right)} \,dt\) \( = \left( {\frac{{3{t^2}}}{2}\hat i + \frac{{3{t^3}}}{3}\hat j} \right) = \left( {\frac{{3{t^2}}}{2}\hat i + {t^3}\hat j} \right)m/s\) Power developed by the force at time ' \(t\) ' \(P=\vec{F} \cdot \vec{v}=\left(6 t \hat{i}+6 t^{2} \hat{j}\right) \cdot\left(\dfrac{3 t^{2}}{2} \hat{i}+t^{3} \hat{j}\right)\) \(P=\left(9 t^{3}+6 t^{5}\right) W\)
JEE - 2024
PHXI06:WORK ENERGY AND POWER
355522
A car of mass \(m\) starts from rest and accelerates so that instantaneous power delivered to the car has a constant magnitude \(P_{0}\). The instantaneous velocity of this car is proportional to
1 \(\dfrac{t}{\sqrt{m}}\)
2 \(t^{2} P_{0}\)
3 \(t^{1 / 2}\)
4 \(t^{-1 / 2}\)
Explanation:
Constant power of a car \(P_{0}=F \cdot v=\) ma. \(v\) \(\begin{gathered}P_{0}=m \dfrac{d v}{d t} \cdot v \Rightarrow P_{0} d t=m v d v \Rightarrow P_{0} . t=\dfrac{m v^{2}}{2} \\v=\sqrt{\dfrac{2 P_{0} t}{m}} \Rightarrow v \alpha \sqrt{t} .\end{gathered}\)
355518
A block of mass \(m\) is allowed to slide down a fixed smooth inclined plane of angle \(\theta\) and length \(l\). What is the magnitude of power developed by the gravitational force when the block reaches the bottom?
1 \(\sqrt{2 m^{2} l(g \sin \theta)^{3}}\)
2 \(\sqrt{(2 / 3) m^{3} l \mathrm{~g}^{2} \sin \theta}\)
3 \(\sqrt{(2 / 3) m^{2} l^{2} g \cos \theta}\)
4 \(\sqrt{(1 / 3) m^{3} l \mathrm{~g}^{2} \sin \theta}\)
Explanation:
The force that accelerates the block is \(m g \sin \theta \Rightarrow a=g \sin \theta\) Velocity of the block after travelling distance \(l\) is \(v^{2}=2 g \sin \theta l\) The instantaneous power is \(\begin{aligned}& P=F v=m g \sin \theta \sqrt{2 g \sin \theta l} \\& P=\sqrt{2 m^{2} g^{3} \sin ^{3} \theta l}\end{aligned}\)
PHXI06:WORK ENERGY AND POWER
355519
A particle moves with a velocity \(5\hat i - 3\hat j + 6\hat k\) \(m{\text{/}}s\) under the influence of a constant force \(F = 10\hat i + 10\hat j + 20\hat k\) \(N\). The instantaneous power applied to the particle is
355520
An engine exerts a force \(F = (20\widehat i - 3\widehat j + 5\widehat k)N\) and moves with velocity \(v = (6\hat i + 20\hat j - 3\hat k)m/s\). The power of the engine ( in watt) is
1 45
2 75
3 20
4 10
Explanation:
Here, Force, \(\vec F = (20\hat i - 3\hat j + 5\hat k)N\) Velocity, \(\vec v = (6\hat i + 20\hat j - 3\hat k)m/s\) \(P = \vec F \cdot \vec v = (20\hat i - 3\hat j + 5\hat k) \cdot (6\hat i + 20\hat j - 3\hat k)\) \(120 - 60 - 15 = 45W\)
PHXI06:WORK ENERGY AND POWER
355521
A body of mass \(2\,kg\) begins to move under the action of a time dependent force given by \(\vec{F}=\left(6 t \hat{i}+6 t^{2} \hat{j}\right) N\). The power developed by the force at this time \(t\) is given by
1 \(\left(3 t^{3}+6 t^{5}\right) W\)
2 \(\left(6 t^{4}+9 t^{5}\right) W\)
3 \(\left(9 t^{5}+6 t^{3}\right) W\)
4 \(\left(9 t^{3}+6 t^{5}\right) W\)
Explanation:
Here, \(\vec F = \left( {6t\hat i + 6{t^2}\hat j} \right)N;\,\,m = 2\;kg\) Acceleration of the body, \(\vec{a}=\dfrac{\vec{F}}{m}\) \(\vec a = \frac{{\left( {6t\hat i + 6{t^2}\hat j} \right)}}{2} = \left( {3t\hat i + 3{t^2}j} \right)m/{s^2}\) Velocity of the body at time \((t)\) is, \(\vec v = \int {\vec a} dt = \int {\left( {3t\hat i + 3{t^2}\hat j} \right)} \,dt\) \( = \left( {\frac{{3{t^2}}}{2}\hat i + \frac{{3{t^3}}}{3}\hat j} \right) = \left( {\frac{{3{t^2}}}{2}\hat i + {t^3}\hat j} \right)m/s\) Power developed by the force at time ' \(t\) ' \(P=\vec{F} \cdot \vec{v}=\left(6 t \hat{i}+6 t^{2} \hat{j}\right) \cdot\left(\dfrac{3 t^{2}}{2} \hat{i}+t^{3} \hat{j}\right)\) \(P=\left(9 t^{3}+6 t^{5}\right) W\)
JEE - 2024
PHXI06:WORK ENERGY AND POWER
355522
A car of mass \(m\) starts from rest and accelerates so that instantaneous power delivered to the car has a constant magnitude \(P_{0}\). The instantaneous velocity of this car is proportional to
1 \(\dfrac{t}{\sqrt{m}}\)
2 \(t^{2} P_{0}\)
3 \(t^{1 / 2}\)
4 \(t^{-1 / 2}\)
Explanation:
Constant power of a car \(P_{0}=F \cdot v=\) ma. \(v\) \(\begin{gathered}P_{0}=m \dfrac{d v}{d t} \cdot v \Rightarrow P_{0} d t=m v d v \Rightarrow P_{0} . t=\dfrac{m v^{2}}{2} \\v=\sqrt{\dfrac{2 P_{0} t}{m}} \Rightarrow v \alpha \sqrt{t} .\end{gathered}\)
