148714
The work done by a force acting on a body is as shown in the graph. The total work done in covering an initial distance of $20 \mathrm{~m}$ is
1 $225 \mathrm{~J}$
2 $200 \mathrm{~J}$
3 $400 \mathrm{~J}$
4 $175 \mathrm{~J}$ $\mathrm{S}($ in $\mathrm{m})$
Explanation:
B Work done is equal to the area under the graph$\mathrm{W}=$ area of $\mathrm{ABC}+$ Area of $\mathrm{BFGC}+$ area of DEF $=\left(\frac{1}{2} \times 5 \times 10\right)+(15 \times 10)+\left(\frac{1}{2} \times 5 \times 10\right)$ $=25+150+25$ $=200 \mathrm{~J}$
JIPMER-2012
Work, Energy and Power
148715
A body of mass $3 \mathrm{~kg}$ in under a force which caused displacement in it, given by $s=\frac{t^{2}}{3}$ in meter with time $\mathbf{t}$ in seconds. What is the work done by the force between time $\mathbf{t}=\mathbf{0}$ and $\mathbf{t}=\mathbf{2}$ is
148716
A particle of mass $100 \mathrm{~g}$ is thrown vertically upwards with a speed of $5 \mathrm{~m} / \mathrm{s}$. The work done by the force of gravity during the time the particle goes up is
1 $-0.5 \mathrm{~J}$
2 $-1.25 \mathrm{~J}$
3 $1.25 \mathrm{~J}$
4 $0.5 \mathrm{~J}$
Explanation:
B Given, mass $(\mathrm{m})=100 \mathrm{~g}=0.1 \mathrm{~kg}, \mathrm{u}=5 \mathrm{~m} / \mathrm{s}$ $\mathrm{v}=0 \mathrm{~m} / \mathrm{s}$ From third equation of motion- $\mathrm{v}^{2}=\mathrm{u}^{2}+2 \mathrm{gh}$ $0=\mathrm{u}^{2}-2 \mathrm{gh}$ $\mathrm{u}^{2}=2 \mathrm{gh}$ $\mathrm{h}=\frac{\mathrm{u}^{2}}{2 \mathrm{~g}}=\frac{(5)^{2}}{2 \times 9.8}=\frac{25}{2 \times 9.8}$ Work done by gravity $(\mathrm{W})=\mathrm{mgh} \cos \theta$ $\mathrm{W}=0.1 \times 9.8 \times \frac{25}{2 \times 9.8} \cos 180^{\circ} \quad\left(\because \theta=180^{\circ}\right)$ $\therefore \mathrm{W} =-0.1 \times \frac{25}{2}=-1.25 \mathrm{~J}$
JCECE-2008
Work, Energy and Power
148717
An engine pumps up $100 \mathrm{~kg}$ of water through a height of $10 \mathrm{~m}$ in $5 \mathrm{~s}$. Given that the efficiency of engine is $60 \%$. If $\mathrm{g}=10 \mathrm{~ms}^{-2}$, the power of the engine is-
1 $3.3 \mathrm{~kW}$
2 $0.33 \mathrm{~kW}$
3 $0.033 \mathrm{~kW}$
4 $33 \mathrm{~kW}$
Explanation:
A Given, $\mathrm{m}=100 \mathrm{~kg}, \mathrm{~h}=10 \mathrm{~m}, \mathrm{t}=5 \mathrm{sec}, \mathrm{g}=10 \mathrm{~ms}^{-2}$ Efficiency $(\eta)=60 \%=0.6$ Work done $=\mathrm{mgh}$ $\mathrm{W} =100 \times 10 \times 10$ $=10000 \mathrm{~J}$ Effective power $=\frac{\text { Work }}{\text { Time }}$ $=\frac{10000}{5}$ $=2000\text { watt }$ Effective power $=$ Efficiency $\times$ Total power $\text { Total power }=\frac{\text { Effective power }}{\text { Efficiency }}$ Total power $=\frac{2000}{0.6}$ $=3333 \text { watt }=3.33 \mathrm{~kW}$
148714
The work done by a force acting on a body is as shown in the graph. The total work done in covering an initial distance of $20 \mathrm{~m}$ is
1 $225 \mathrm{~J}$
2 $200 \mathrm{~J}$
3 $400 \mathrm{~J}$
4 $175 \mathrm{~J}$ $\mathrm{S}($ in $\mathrm{m})$
Explanation:
B Work done is equal to the area under the graph$\mathrm{W}=$ area of $\mathrm{ABC}+$ Area of $\mathrm{BFGC}+$ area of DEF $=\left(\frac{1}{2} \times 5 \times 10\right)+(15 \times 10)+\left(\frac{1}{2} \times 5 \times 10\right)$ $=25+150+25$ $=200 \mathrm{~J}$
JIPMER-2012
Work, Energy and Power
148715
A body of mass $3 \mathrm{~kg}$ in under a force which caused displacement in it, given by $s=\frac{t^{2}}{3}$ in meter with time $\mathbf{t}$ in seconds. What is the work done by the force between time $\mathbf{t}=\mathbf{0}$ and $\mathbf{t}=\mathbf{2}$ is
148716
A particle of mass $100 \mathrm{~g}$ is thrown vertically upwards with a speed of $5 \mathrm{~m} / \mathrm{s}$. The work done by the force of gravity during the time the particle goes up is
1 $-0.5 \mathrm{~J}$
2 $-1.25 \mathrm{~J}$
3 $1.25 \mathrm{~J}$
4 $0.5 \mathrm{~J}$
Explanation:
B Given, mass $(\mathrm{m})=100 \mathrm{~g}=0.1 \mathrm{~kg}, \mathrm{u}=5 \mathrm{~m} / \mathrm{s}$ $\mathrm{v}=0 \mathrm{~m} / \mathrm{s}$ From third equation of motion- $\mathrm{v}^{2}=\mathrm{u}^{2}+2 \mathrm{gh}$ $0=\mathrm{u}^{2}-2 \mathrm{gh}$ $\mathrm{u}^{2}=2 \mathrm{gh}$ $\mathrm{h}=\frac{\mathrm{u}^{2}}{2 \mathrm{~g}}=\frac{(5)^{2}}{2 \times 9.8}=\frac{25}{2 \times 9.8}$ Work done by gravity $(\mathrm{W})=\mathrm{mgh} \cos \theta$ $\mathrm{W}=0.1 \times 9.8 \times \frac{25}{2 \times 9.8} \cos 180^{\circ} \quad\left(\because \theta=180^{\circ}\right)$ $\therefore \mathrm{W} =-0.1 \times \frac{25}{2}=-1.25 \mathrm{~J}$
JCECE-2008
Work, Energy and Power
148717
An engine pumps up $100 \mathrm{~kg}$ of water through a height of $10 \mathrm{~m}$ in $5 \mathrm{~s}$. Given that the efficiency of engine is $60 \%$. If $\mathrm{g}=10 \mathrm{~ms}^{-2}$, the power of the engine is-
1 $3.3 \mathrm{~kW}$
2 $0.33 \mathrm{~kW}$
3 $0.033 \mathrm{~kW}$
4 $33 \mathrm{~kW}$
Explanation:
A Given, $\mathrm{m}=100 \mathrm{~kg}, \mathrm{~h}=10 \mathrm{~m}, \mathrm{t}=5 \mathrm{sec}, \mathrm{g}=10 \mathrm{~ms}^{-2}$ Efficiency $(\eta)=60 \%=0.6$ Work done $=\mathrm{mgh}$ $\mathrm{W} =100 \times 10 \times 10$ $=10000 \mathrm{~J}$ Effective power $=\frac{\text { Work }}{\text { Time }}$ $=\frac{10000}{5}$ $=2000\text { watt }$ Effective power $=$ Efficiency $\times$ Total power $\text { Total power }=\frac{\text { Effective power }}{\text { Efficiency }}$ Total power $=\frac{2000}{0.6}$ $=3333 \text { watt }=3.33 \mathrm{~kW}$
148714
The work done by a force acting on a body is as shown in the graph. The total work done in covering an initial distance of $20 \mathrm{~m}$ is
1 $225 \mathrm{~J}$
2 $200 \mathrm{~J}$
3 $400 \mathrm{~J}$
4 $175 \mathrm{~J}$ $\mathrm{S}($ in $\mathrm{m})$
Explanation:
B Work done is equal to the area under the graph$\mathrm{W}=$ area of $\mathrm{ABC}+$ Area of $\mathrm{BFGC}+$ area of DEF $=\left(\frac{1}{2} \times 5 \times 10\right)+(15 \times 10)+\left(\frac{1}{2} \times 5 \times 10\right)$ $=25+150+25$ $=200 \mathrm{~J}$
JIPMER-2012
Work, Energy and Power
148715
A body of mass $3 \mathrm{~kg}$ in under a force which caused displacement in it, given by $s=\frac{t^{2}}{3}$ in meter with time $\mathbf{t}$ in seconds. What is the work done by the force between time $\mathbf{t}=\mathbf{0}$ and $\mathbf{t}=\mathbf{2}$ is
148716
A particle of mass $100 \mathrm{~g}$ is thrown vertically upwards with a speed of $5 \mathrm{~m} / \mathrm{s}$. The work done by the force of gravity during the time the particle goes up is
1 $-0.5 \mathrm{~J}$
2 $-1.25 \mathrm{~J}$
3 $1.25 \mathrm{~J}$
4 $0.5 \mathrm{~J}$
Explanation:
B Given, mass $(\mathrm{m})=100 \mathrm{~g}=0.1 \mathrm{~kg}, \mathrm{u}=5 \mathrm{~m} / \mathrm{s}$ $\mathrm{v}=0 \mathrm{~m} / \mathrm{s}$ From third equation of motion- $\mathrm{v}^{2}=\mathrm{u}^{2}+2 \mathrm{gh}$ $0=\mathrm{u}^{2}-2 \mathrm{gh}$ $\mathrm{u}^{2}=2 \mathrm{gh}$ $\mathrm{h}=\frac{\mathrm{u}^{2}}{2 \mathrm{~g}}=\frac{(5)^{2}}{2 \times 9.8}=\frac{25}{2 \times 9.8}$ Work done by gravity $(\mathrm{W})=\mathrm{mgh} \cos \theta$ $\mathrm{W}=0.1 \times 9.8 \times \frac{25}{2 \times 9.8} \cos 180^{\circ} \quad\left(\because \theta=180^{\circ}\right)$ $\therefore \mathrm{W} =-0.1 \times \frac{25}{2}=-1.25 \mathrm{~J}$
JCECE-2008
Work, Energy and Power
148717
An engine pumps up $100 \mathrm{~kg}$ of water through a height of $10 \mathrm{~m}$ in $5 \mathrm{~s}$. Given that the efficiency of engine is $60 \%$. If $\mathrm{g}=10 \mathrm{~ms}^{-2}$, the power of the engine is-
1 $3.3 \mathrm{~kW}$
2 $0.33 \mathrm{~kW}$
3 $0.033 \mathrm{~kW}$
4 $33 \mathrm{~kW}$
Explanation:
A Given, $\mathrm{m}=100 \mathrm{~kg}, \mathrm{~h}=10 \mathrm{~m}, \mathrm{t}=5 \mathrm{sec}, \mathrm{g}=10 \mathrm{~ms}^{-2}$ Efficiency $(\eta)=60 \%=0.6$ Work done $=\mathrm{mgh}$ $\mathrm{W} =100 \times 10 \times 10$ $=10000 \mathrm{~J}$ Effective power $=\frac{\text { Work }}{\text { Time }}$ $=\frac{10000}{5}$ $=2000\text { watt }$ Effective power $=$ Efficiency $\times$ Total power $\text { Total power }=\frac{\text { Effective power }}{\text { Efficiency }}$ Total power $=\frac{2000}{0.6}$ $=3333 \text { watt }=3.33 \mathrm{~kW}$
148714
The work done by a force acting on a body is as shown in the graph. The total work done in covering an initial distance of $20 \mathrm{~m}$ is
1 $225 \mathrm{~J}$
2 $200 \mathrm{~J}$
3 $400 \mathrm{~J}$
4 $175 \mathrm{~J}$ $\mathrm{S}($ in $\mathrm{m})$
Explanation:
B Work done is equal to the area under the graph$\mathrm{W}=$ area of $\mathrm{ABC}+$ Area of $\mathrm{BFGC}+$ area of DEF $=\left(\frac{1}{2} \times 5 \times 10\right)+(15 \times 10)+\left(\frac{1}{2} \times 5 \times 10\right)$ $=25+150+25$ $=200 \mathrm{~J}$
JIPMER-2012
Work, Energy and Power
148715
A body of mass $3 \mathrm{~kg}$ in under a force which caused displacement in it, given by $s=\frac{t^{2}}{3}$ in meter with time $\mathbf{t}$ in seconds. What is the work done by the force between time $\mathbf{t}=\mathbf{0}$ and $\mathbf{t}=\mathbf{2}$ is
148716
A particle of mass $100 \mathrm{~g}$ is thrown vertically upwards with a speed of $5 \mathrm{~m} / \mathrm{s}$. The work done by the force of gravity during the time the particle goes up is
1 $-0.5 \mathrm{~J}$
2 $-1.25 \mathrm{~J}$
3 $1.25 \mathrm{~J}$
4 $0.5 \mathrm{~J}$
Explanation:
B Given, mass $(\mathrm{m})=100 \mathrm{~g}=0.1 \mathrm{~kg}, \mathrm{u}=5 \mathrm{~m} / \mathrm{s}$ $\mathrm{v}=0 \mathrm{~m} / \mathrm{s}$ From third equation of motion- $\mathrm{v}^{2}=\mathrm{u}^{2}+2 \mathrm{gh}$ $0=\mathrm{u}^{2}-2 \mathrm{gh}$ $\mathrm{u}^{2}=2 \mathrm{gh}$ $\mathrm{h}=\frac{\mathrm{u}^{2}}{2 \mathrm{~g}}=\frac{(5)^{2}}{2 \times 9.8}=\frac{25}{2 \times 9.8}$ Work done by gravity $(\mathrm{W})=\mathrm{mgh} \cos \theta$ $\mathrm{W}=0.1 \times 9.8 \times \frac{25}{2 \times 9.8} \cos 180^{\circ} \quad\left(\because \theta=180^{\circ}\right)$ $\therefore \mathrm{W} =-0.1 \times \frac{25}{2}=-1.25 \mathrm{~J}$
JCECE-2008
Work, Energy and Power
148717
An engine pumps up $100 \mathrm{~kg}$ of water through a height of $10 \mathrm{~m}$ in $5 \mathrm{~s}$. Given that the efficiency of engine is $60 \%$. If $\mathrm{g}=10 \mathrm{~ms}^{-2}$, the power of the engine is-
1 $3.3 \mathrm{~kW}$
2 $0.33 \mathrm{~kW}$
3 $0.033 \mathrm{~kW}$
4 $33 \mathrm{~kW}$
Explanation:
A Given, $\mathrm{m}=100 \mathrm{~kg}, \mathrm{~h}=10 \mathrm{~m}, \mathrm{t}=5 \mathrm{sec}, \mathrm{g}=10 \mathrm{~ms}^{-2}$ Efficiency $(\eta)=60 \%=0.6$ Work done $=\mathrm{mgh}$ $\mathrm{W} =100 \times 10 \times 10$ $=10000 \mathrm{~J}$ Effective power $=\frac{\text { Work }}{\text { Time }}$ $=\frac{10000}{5}$ $=2000\text { watt }$ Effective power $=$ Efficiency $\times$ Total power $\text { Total power }=\frac{\text { Effective power }}{\text { Efficiency }}$ Total power $=\frac{2000}{0.6}$ $=3333 \text { watt }=3.33 \mathrm{~kW}$
