355471
A 60 \(HP\) electric motor lifts an elevator having a maximum total load capacity of 2000 \(kg\). If the frictional force on the elevator is 4000 \(N\), the speed of the elevator at full load is close to: \(\left( {1HP = 746\;W,\;g = 10\;m{s^{ - 2}}} \right)\)
1 \(1.7\;m{s^{ - 1}}\)
2 \(1.9\;m{s^{ - 1}}\)
3 \(2.0\;m{s^{ - 1}}\)
4 \(1.5\;m{s^{ - 1}}\)
Explanation:
\(P=f v+M g v\) Applied power \(=4000 \times v+20000 v\) \(60 \times 746 = 4000v + 20000v\) \(v \approx 1.9m/s\)
JEE - 2020
PHXI06:WORK ENERGY AND POWER
355472
An object of mass \(m\) moves horizontally, increasing in speed from 0 to \(v\) in a time \(t\). The power necessary to accelerate the object during this time period is
1 \(\dfrac{m v^{2} t}{2}\)
2 \(\dfrac{m v^{2}}{2}\)
3 \(2 m v^{2}\)
4 \(\dfrac{m v^{2}}{2 t}\)
Explanation:
Power, \(P=\dfrac{W}{t}\) Since, \(\quad K_{i}=\) initial \(K E=0\) \(K_{f}=\) final \(K E=\dfrac{1}{2} m v^{2}\) From work-energy theorem, Work done \(=\) Change in \(\mathrm{KE}\) \(\begin{aligned}& \therefore \quad P=\dfrac{K_{f}-K_{i}}{t}=\dfrac{\dfrac{1}{2} m v^{2}-0}{t} \\& \Rightarrow \quad P=\dfrac{m v^{2}}{2 t}\end{aligned}\)
PHXI06:WORK ENERGY AND POWER
355473
Find the speed \({v}\) acquired by a car of mass \({m}\), driven with constant power \({P}\), and covers a distance \({x}\) traveled from origin.
1 \({\left(\dfrac{3 x P}{m}\right)^{\frac{1}{3}}}\)
2 \({\left(\dfrac{2 x P}{m}\right)^{\frac{1}{2}}}\)
3 \({\left(\dfrac{x P}{m}\right)^{\frac{1}{3}}}\)
4 \({\left(\dfrac{2 x P}{3 m}\right)^{\frac{1}{4}}}\)
Explanation:
\({P=F v}\) \({P=mav}\) \({P=m v^{2} \dfrac{d v}{d x} \quad\left[\because a=\dfrac{v d v}{d x}\right]}\) \({\int_{0}^{v} v^{2} d v=\dfrac{P}{m} \int_{0}^{x} d x}\) \({\dfrac{v^{3}}{3}=\dfrac{P}{m} x}\) \(v=\left(\dfrac{3 x P}{m}\right)^{\frac{1}{3}}\)
PHXI06:WORK ENERGY AND POWER
355474
The machine gun fires 240 bullets per minute. If the mass of each bullet is \(10\;g\) and the velocity of the bullets is \(600\;m{s^{ - 1}},\) the power (in \(k W\) ) of the gun is
1 43200
2 432
3 72
4 7.2
Explanation:
Given, \(n=240,=4\) bullets per second, \(m = 10g = 10 \times {10^{ - 3}}\;kg,\) \(t = 1\,s\) and \(v = 600\;m{s^{ - 1}}\) Work done by the gun \(=\) Total kinetic energy of the bullets \(=n \dfrac{1}{2} m v^{2}=4 \times \dfrac{1}{2} \times 10 \times 10^{-3} \times(600)^{2}\) \(=2 \times 10 \times 10^{-3} \times 600 \times 600\) \(\therefore \quad\) Power of gun \(=\dfrac{\text { Work done }}{\text { Time taken }}\) \(=\dfrac{2 \times 10 \times 10^{-3} \times 600 \times 600}{1}\) \( = 7.2\;k\,W\)
PHXI06:WORK ENERGY AND POWER
355475
An engine develops 10 \(kW\) of power. how much time will it take to lift a mass of 200 \(kg\) to a height of \(40\;m\left( {\;g = 10\;m/{s^2}} \right)\)
355471
A 60 \(HP\) electric motor lifts an elevator having a maximum total load capacity of 2000 \(kg\). If the frictional force on the elevator is 4000 \(N\), the speed of the elevator at full load is close to: \(\left( {1HP = 746\;W,\;g = 10\;m{s^{ - 2}}} \right)\)
1 \(1.7\;m{s^{ - 1}}\)
2 \(1.9\;m{s^{ - 1}}\)
3 \(2.0\;m{s^{ - 1}}\)
4 \(1.5\;m{s^{ - 1}}\)
Explanation:
\(P=f v+M g v\) Applied power \(=4000 \times v+20000 v\) \(60 \times 746 = 4000v + 20000v\) \(v \approx 1.9m/s\)
JEE - 2020
PHXI06:WORK ENERGY AND POWER
355472
An object of mass \(m\) moves horizontally, increasing in speed from 0 to \(v\) in a time \(t\). The power necessary to accelerate the object during this time period is
1 \(\dfrac{m v^{2} t}{2}\)
2 \(\dfrac{m v^{2}}{2}\)
3 \(2 m v^{2}\)
4 \(\dfrac{m v^{2}}{2 t}\)
Explanation:
Power, \(P=\dfrac{W}{t}\) Since, \(\quad K_{i}=\) initial \(K E=0\) \(K_{f}=\) final \(K E=\dfrac{1}{2} m v^{2}\) From work-energy theorem, Work done \(=\) Change in \(\mathrm{KE}\) \(\begin{aligned}& \therefore \quad P=\dfrac{K_{f}-K_{i}}{t}=\dfrac{\dfrac{1}{2} m v^{2}-0}{t} \\& \Rightarrow \quad P=\dfrac{m v^{2}}{2 t}\end{aligned}\)
PHXI06:WORK ENERGY AND POWER
355473
Find the speed \({v}\) acquired by a car of mass \({m}\), driven with constant power \({P}\), and covers a distance \({x}\) traveled from origin.
1 \({\left(\dfrac{3 x P}{m}\right)^{\frac{1}{3}}}\)
2 \({\left(\dfrac{2 x P}{m}\right)^{\frac{1}{2}}}\)
3 \({\left(\dfrac{x P}{m}\right)^{\frac{1}{3}}}\)
4 \({\left(\dfrac{2 x P}{3 m}\right)^{\frac{1}{4}}}\)
Explanation:
\({P=F v}\) \({P=mav}\) \({P=m v^{2} \dfrac{d v}{d x} \quad\left[\because a=\dfrac{v d v}{d x}\right]}\) \({\int_{0}^{v} v^{2} d v=\dfrac{P}{m} \int_{0}^{x} d x}\) \({\dfrac{v^{3}}{3}=\dfrac{P}{m} x}\) \(v=\left(\dfrac{3 x P}{m}\right)^{\frac{1}{3}}\)
PHXI06:WORK ENERGY AND POWER
355474
The machine gun fires 240 bullets per minute. If the mass of each bullet is \(10\;g\) and the velocity of the bullets is \(600\;m{s^{ - 1}},\) the power (in \(k W\) ) of the gun is
1 43200
2 432
3 72
4 7.2
Explanation:
Given, \(n=240,=4\) bullets per second, \(m = 10g = 10 \times {10^{ - 3}}\;kg,\) \(t = 1\,s\) and \(v = 600\;m{s^{ - 1}}\) Work done by the gun \(=\) Total kinetic energy of the bullets \(=n \dfrac{1}{2} m v^{2}=4 \times \dfrac{1}{2} \times 10 \times 10^{-3} \times(600)^{2}\) \(=2 \times 10 \times 10^{-3} \times 600 \times 600\) \(\therefore \quad\) Power of gun \(=\dfrac{\text { Work done }}{\text { Time taken }}\) \(=\dfrac{2 \times 10 \times 10^{-3} \times 600 \times 600}{1}\) \( = 7.2\;k\,W\)
PHXI06:WORK ENERGY AND POWER
355475
An engine develops 10 \(kW\) of power. how much time will it take to lift a mass of 200 \(kg\) to a height of \(40\;m\left( {\;g = 10\;m/{s^2}} \right)\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXI06:WORK ENERGY AND POWER
355471
A 60 \(HP\) electric motor lifts an elevator having a maximum total load capacity of 2000 \(kg\). If the frictional force on the elevator is 4000 \(N\), the speed of the elevator at full load is close to: \(\left( {1HP = 746\;W,\;g = 10\;m{s^{ - 2}}} \right)\)
1 \(1.7\;m{s^{ - 1}}\)
2 \(1.9\;m{s^{ - 1}}\)
3 \(2.0\;m{s^{ - 1}}\)
4 \(1.5\;m{s^{ - 1}}\)
Explanation:
\(P=f v+M g v\) Applied power \(=4000 \times v+20000 v\) \(60 \times 746 = 4000v + 20000v\) \(v \approx 1.9m/s\)
JEE - 2020
PHXI06:WORK ENERGY AND POWER
355472
An object of mass \(m\) moves horizontally, increasing in speed from 0 to \(v\) in a time \(t\). The power necessary to accelerate the object during this time period is
1 \(\dfrac{m v^{2} t}{2}\)
2 \(\dfrac{m v^{2}}{2}\)
3 \(2 m v^{2}\)
4 \(\dfrac{m v^{2}}{2 t}\)
Explanation:
Power, \(P=\dfrac{W}{t}\) Since, \(\quad K_{i}=\) initial \(K E=0\) \(K_{f}=\) final \(K E=\dfrac{1}{2} m v^{2}\) From work-energy theorem, Work done \(=\) Change in \(\mathrm{KE}\) \(\begin{aligned}& \therefore \quad P=\dfrac{K_{f}-K_{i}}{t}=\dfrac{\dfrac{1}{2} m v^{2}-0}{t} \\& \Rightarrow \quad P=\dfrac{m v^{2}}{2 t}\end{aligned}\)
PHXI06:WORK ENERGY AND POWER
355473
Find the speed \({v}\) acquired by a car of mass \({m}\), driven with constant power \({P}\), and covers a distance \({x}\) traveled from origin.
1 \({\left(\dfrac{3 x P}{m}\right)^{\frac{1}{3}}}\)
2 \({\left(\dfrac{2 x P}{m}\right)^{\frac{1}{2}}}\)
3 \({\left(\dfrac{x P}{m}\right)^{\frac{1}{3}}}\)
4 \({\left(\dfrac{2 x P}{3 m}\right)^{\frac{1}{4}}}\)
Explanation:
\({P=F v}\) \({P=mav}\) \({P=m v^{2} \dfrac{d v}{d x} \quad\left[\because a=\dfrac{v d v}{d x}\right]}\) \({\int_{0}^{v} v^{2} d v=\dfrac{P}{m} \int_{0}^{x} d x}\) \({\dfrac{v^{3}}{3}=\dfrac{P}{m} x}\) \(v=\left(\dfrac{3 x P}{m}\right)^{\frac{1}{3}}\)
PHXI06:WORK ENERGY AND POWER
355474
The machine gun fires 240 bullets per minute. If the mass of each bullet is \(10\;g\) and the velocity of the bullets is \(600\;m{s^{ - 1}},\) the power (in \(k W\) ) of the gun is
1 43200
2 432
3 72
4 7.2
Explanation:
Given, \(n=240,=4\) bullets per second, \(m = 10g = 10 \times {10^{ - 3}}\;kg,\) \(t = 1\,s\) and \(v = 600\;m{s^{ - 1}}\) Work done by the gun \(=\) Total kinetic energy of the bullets \(=n \dfrac{1}{2} m v^{2}=4 \times \dfrac{1}{2} \times 10 \times 10^{-3} \times(600)^{2}\) \(=2 \times 10 \times 10^{-3} \times 600 \times 600\) \(\therefore \quad\) Power of gun \(=\dfrac{\text { Work done }}{\text { Time taken }}\) \(=\dfrac{2 \times 10 \times 10^{-3} \times 600 \times 600}{1}\) \( = 7.2\;k\,W\)
PHXI06:WORK ENERGY AND POWER
355475
An engine develops 10 \(kW\) of power. how much time will it take to lift a mass of 200 \(kg\) to a height of \(40\;m\left( {\;g = 10\;m/{s^2}} \right)\)
355471
A 60 \(HP\) electric motor lifts an elevator having a maximum total load capacity of 2000 \(kg\). If the frictional force on the elevator is 4000 \(N\), the speed of the elevator at full load is close to: \(\left( {1HP = 746\;W,\;g = 10\;m{s^{ - 2}}} \right)\)
1 \(1.7\;m{s^{ - 1}}\)
2 \(1.9\;m{s^{ - 1}}\)
3 \(2.0\;m{s^{ - 1}}\)
4 \(1.5\;m{s^{ - 1}}\)
Explanation:
\(P=f v+M g v\) Applied power \(=4000 \times v+20000 v\) \(60 \times 746 = 4000v + 20000v\) \(v \approx 1.9m/s\)
JEE - 2020
PHXI06:WORK ENERGY AND POWER
355472
An object of mass \(m\) moves horizontally, increasing in speed from 0 to \(v\) in a time \(t\). The power necessary to accelerate the object during this time period is
1 \(\dfrac{m v^{2} t}{2}\)
2 \(\dfrac{m v^{2}}{2}\)
3 \(2 m v^{2}\)
4 \(\dfrac{m v^{2}}{2 t}\)
Explanation:
Power, \(P=\dfrac{W}{t}\) Since, \(\quad K_{i}=\) initial \(K E=0\) \(K_{f}=\) final \(K E=\dfrac{1}{2} m v^{2}\) From work-energy theorem, Work done \(=\) Change in \(\mathrm{KE}\) \(\begin{aligned}& \therefore \quad P=\dfrac{K_{f}-K_{i}}{t}=\dfrac{\dfrac{1}{2} m v^{2}-0}{t} \\& \Rightarrow \quad P=\dfrac{m v^{2}}{2 t}\end{aligned}\)
PHXI06:WORK ENERGY AND POWER
355473
Find the speed \({v}\) acquired by a car of mass \({m}\), driven with constant power \({P}\), and covers a distance \({x}\) traveled from origin.
1 \({\left(\dfrac{3 x P}{m}\right)^{\frac{1}{3}}}\)
2 \({\left(\dfrac{2 x P}{m}\right)^{\frac{1}{2}}}\)
3 \({\left(\dfrac{x P}{m}\right)^{\frac{1}{3}}}\)
4 \({\left(\dfrac{2 x P}{3 m}\right)^{\frac{1}{4}}}\)
Explanation:
\({P=F v}\) \({P=mav}\) \({P=m v^{2} \dfrac{d v}{d x} \quad\left[\because a=\dfrac{v d v}{d x}\right]}\) \({\int_{0}^{v} v^{2} d v=\dfrac{P}{m} \int_{0}^{x} d x}\) \({\dfrac{v^{3}}{3}=\dfrac{P}{m} x}\) \(v=\left(\dfrac{3 x P}{m}\right)^{\frac{1}{3}}\)
PHXI06:WORK ENERGY AND POWER
355474
The machine gun fires 240 bullets per minute. If the mass of each bullet is \(10\;g\) and the velocity of the bullets is \(600\;m{s^{ - 1}},\) the power (in \(k W\) ) of the gun is
1 43200
2 432
3 72
4 7.2
Explanation:
Given, \(n=240,=4\) bullets per second, \(m = 10g = 10 \times {10^{ - 3}}\;kg,\) \(t = 1\,s\) and \(v = 600\;m{s^{ - 1}}\) Work done by the gun \(=\) Total kinetic energy of the bullets \(=n \dfrac{1}{2} m v^{2}=4 \times \dfrac{1}{2} \times 10 \times 10^{-3} \times(600)^{2}\) \(=2 \times 10 \times 10^{-3} \times 600 \times 600\) \(\therefore \quad\) Power of gun \(=\dfrac{\text { Work done }}{\text { Time taken }}\) \(=\dfrac{2 \times 10 \times 10^{-3} \times 600 \times 600}{1}\) \( = 7.2\;k\,W\)
PHXI06:WORK ENERGY AND POWER
355475
An engine develops 10 \(kW\) of power. how much time will it take to lift a mass of 200 \(kg\) to a height of \(40\;m\left( {\;g = 10\;m/{s^2}} \right)\)
355471
A 60 \(HP\) electric motor lifts an elevator having a maximum total load capacity of 2000 \(kg\). If the frictional force on the elevator is 4000 \(N\), the speed of the elevator at full load is close to: \(\left( {1HP = 746\;W,\;g = 10\;m{s^{ - 2}}} \right)\)
1 \(1.7\;m{s^{ - 1}}\)
2 \(1.9\;m{s^{ - 1}}\)
3 \(2.0\;m{s^{ - 1}}\)
4 \(1.5\;m{s^{ - 1}}\)
Explanation:
\(P=f v+M g v\) Applied power \(=4000 \times v+20000 v\) \(60 \times 746 = 4000v + 20000v\) \(v \approx 1.9m/s\)
JEE - 2020
PHXI06:WORK ENERGY AND POWER
355472
An object of mass \(m\) moves horizontally, increasing in speed from 0 to \(v\) in a time \(t\). The power necessary to accelerate the object during this time period is
1 \(\dfrac{m v^{2} t}{2}\)
2 \(\dfrac{m v^{2}}{2}\)
3 \(2 m v^{2}\)
4 \(\dfrac{m v^{2}}{2 t}\)
Explanation:
Power, \(P=\dfrac{W}{t}\) Since, \(\quad K_{i}=\) initial \(K E=0\) \(K_{f}=\) final \(K E=\dfrac{1}{2} m v^{2}\) From work-energy theorem, Work done \(=\) Change in \(\mathrm{KE}\) \(\begin{aligned}& \therefore \quad P=\dfrac{K_{f}-K_{i}}{t}=\dfrac{\dfrac{1}{2} m v^{2}-0}{t} \\& \Rightarrow \quad P=\dfrac{m v^{2}}{2 t}\end{aligned}\)
PHXI06:WORK ENERGY AND POWER
355473
Find the speed \({v}\) acquired by a car of mass \({m}\), driven with constant power \({P}\), and covers a distance \({x}\) traveled from origin.
1 \({\left(\dfrac{3 x P}{m}\right)^{\frac{1}{3}}}\)
2 \({\left(\dfrac{2 x P}{m}\right)^{\frac{1}{2}}}\)
3 \({\left(\dfrac{x P}{m}\right)^{\frac{1}{3}}}\)
4 \({\left(\dfrac{2 x P}{3 m}\right)^{\frac{1}{4}}}\)
Explanation:
\({P=F v}\) \({P=mav}\) \({P=m v^{2} \dfrac{d v}{d x} \quad\left[\because a=\dfrac{v d v}{d x}\right]}\) \({\int_{0}^{v} v^{2} d v=\dfrac{P}{m} \int_{0}^{x} d x}\) \({\dfrac{v^{3}}{3}=\dfrac{P}{m} x}\) \(v=\left(\dfrac{3 x P}{m}\right)^{\frac{1}{3}}\)
PHXI06:WORK ENERGY AND POWER
355474
The machine gun fires 240 bullets per minute. If the mass of each bullet is \(10\;g\) and the velocity of the bullets is \(600\;m{s^{ - 1}},\) the power (in \(k W\) ) of the gun is
1 43200
2 432
3 72
4 7.2
Explanation:
Given, \(n=240,=4\) bullets per second, \(m = 10g = 10 \times {10^{ - 3}}\;kg,\) \(t = 1\,s\) and \(v = 600\;m{s^{ - 1}}\) Work done by the gun \(=\) Total kinetic energy of the bullets \(=n \dfrac{1}{2} m v^{2}=4 \times \dfrac{1}{2} \times 10 \times 10^{-3} \times(600)^{2}\) \(=2 \times 10 \times 10^{-3} \times 600 \times 600\) \(\therefore \quad\) Power of gun \(=\dfrac{\text { Work done }}{\text { Time taken }}\) \(=\dfrac{2 \times 10 \times 10^{-3} \times 600 \times 600}{1}\) \( = 7.2\;k\,W\)
PHXI06:WORK ENERGY AND POWER
355475
An engine develops 10 \(kW\) of power. how much time will it take to lift a mass of 200 \(kg\) to a height of \(40\;m\left( {\;g = 10\;m/{s^2}} \right)\)