148952
When a ball of mass $50 \mathrm{~g}$ is thrown upwards. it raises to a maximum height of $100 \mathrm{~m}$. At what height its K.E. will be reduced to $70 \%$
1 $30 \mathrm{~m}$
2 $40 \mathrm{~m}$
3 $60 \mathrm{~m}$
4 $70 \mathrm{~m}$
Explanation:
A Given, Mass $(\mathrm{m})=50 \mathrm{~g}=0.05 \mathrm{~kg}$ Height $(\mathrm{h})=100 \mathrm{~m}$ At maximum height $\mathrm{v}^{2}=\mathrm{u}^{2}-2 \mathrm{gh}$ $0=\mathrm{u}^{2}-2(10)(100)$ $\mathrm{u}=\sqrt{2000}=20 \sqrt{5}$ $\mathrm{KE}=\frac{1}{2} \mathrm{mu}^{2}=\frac{1}{2}(0.05)(20 \sqrt{5})^{2}$ $=\frac{1}{2} \times(0.05)(2000)=50 \mathrm{~J}$ Now, $\mathrm{K} \cdot \mathrm{E}^{\prime} =70 \% \mathrm{~K} \cdot \mathrm{E}$ $\mathrm{K} \cdot \mathrm{E}^{\prime} =0.7 \mathrm{KE}=0.7 \times 50$ $=35 \mathrm{~J}$ Then, $\mathrm{K} . \mathrm{E}^{\prime}=\frac{1}{2} \mathrm{mv}^{2}$ $\mathrm{v}=\sqrt{\frac{35 \times 2}{0.05}}=\sqrt{1400}$ Using the equation of motion, $v^{2}=u^{2}-2 g h$ $1400=2000-2(10) h$ $-600=-20 h$ $h=30 \mathrm{~m}$
Shift-II]
Work, Energy and Power
148953
As shown in the figure below, the force $F$ on a particle varies with position in a certain manner. The kinetic energy of the particle at $\mathbf{x}$ $=0$ is $12 \mathrm{~J}$. What is its kinetic energy at $x=14$ ?
1 $42 \mathrm{~J}$
2 $70 \mathrm{~J}$
3 $40 \mathrm{~J}$
4 $0 \mathrm{~J}$
Explanation:
A Given, Kinetic energy (K.E.) $)_{(\mathrm{x}=0)}=12 \mathrm{~J}$ Work done $(\mathrm{W})=$ Area under $(\mathrm{F}-\mathrm{x})$ graph Work done $(\mathrm{W})=$ Area of $\triangle \mathrm{AOB}+$ Area of $\square \mathrm{BCDE}+$ Area of $\square \mathrm{EFGH}$ $\mathrm{W} =\frac{1}{2} \times 6 \times 10+(-5) \times(10-6)+5 \times(14-10)$ $\mathrm{W} =30-20+20$ $\mathrm{~W} =30 \mathrm{~J}$ We know that Work done $(\mathrm{W})=$ Change in kinetic energy Work done $=(\text { K.E. })_{\mathrm{f}}-(\text { K.E. })_{\mathrm{i}}$ So, $30 \mathrm{~J}=(\mathrm{K} . \mathrm{E} .)_{\mathrm{f}}-12 \mathrm{~J}$ (K.E. $)_{\mathrm{f}}=42 \mathrm{~J}$
Assam CEE-2021
Work, Energy and Power
148954
A particle is moving along the $x$-axis and has potential energy $U=1+12 x+6 x^{2}$. If the particle is now released at $x=-5$, the maximum value of $x$ can be
1 -1
2 0
3 1
4 3
Explanation:
D Given that, Potential energy $(U)=1+12 x+6 x^{2}$ $\because \mathrm{F} =\frac{-\mathrm{dU}}{\mathrm{dx}}$ $=\frac{-\mathrm{d}\left(1+12 \mathrm{x}+6 \mathrm{x}^{2}\right)}{\mathrm{dx}}$ $\mathrm{F} =0-12-12 \mathrm{x}$ At equilibrium position - $F=0$ $-12-12 x=0$ $x=-1$ It will oscillate about $\mathrm{x}=-1$ with amplitude of $4 \mathrm{~m}$ $\therefore$ Maximum value of $x$ will be $3 \mathrm{~m}$
Assam CEE-2021
Work, Energy and Power
148955
A machine gun fires 360 bullets per minute. Each bullet travels with velocity of $500 \mathrm{~m} / \mathrm{s}$. If the power of the machine gun is $4.5 \mathrm{~kW}$ then the mass of each bullet is
1 $2 \mathrm{~g}$
2 $5 \mathrm{~g}$
3 $6 \mathrm{~g}$
4 $10 \mathrm{~g}$
Explanation:
C Given, No. of bullets $(n)=360$, Velocity of each bullet $(\mathrm{v})=500 \mathrm{~m} / \mathrm{s}$ We know, Power $(P)=4500$ watt, $t=1$ minute $=60 \mathrm{~s}$ Kinetic energy of 360 bullets (K.E) $=n \times \frac{1}{2} m v^{2}$ $=360 \times \frac{1}{2} \times \mathrm{m} \times(500)^{2}$ $\mathrm{~K} . \mathrm{E}= \left(45 \times 10^{6} \mathrm{~m}\right) \mathrm{J}$ $\text { Power }(\mathrm{P}) =\frac{\mathrm{K} . \mathrm{E}}{\text { Time }}$ $4500 =\frac{45 \times 10^{6} \times \mathrm{m}}{60}$ $\mathrm{~m} =\frac{4500 \times 60}{45 \times 10^{6}}$ $\mathrm{~m} =0.006 \mathrm{~kg}$ $\mathrm{~m} =6 \mathrm{~g}$
148952
When a ball of mass $50 \mathrm{~g}$ is thrown upwards. it raises to a maximum height of $100 \mathrm{~m}$. At what height its K.E. will be reduced to $70 \%$
1 $30 \mathrm{~m}$
2 $40 \mathrm{~m}$
3 $60 \mathrm{~m}$
4 $70 \mathrm{~m}$
Explanation:
A Given, Mass $(\mathrm{m})=50 \mathrm{~g}=0.05 \mathrm{~kg}$ Height $(\mathrm{h})=100 \mathrm{~m}$ At maximum height $\mathrm{v}^{2}=\mathrm{u}^{2}-2 \mathrm{gh}$ $0=\mathrm{u}^{2}-2(10)(100)$ $\mathrm{u}=\sqrt{2000}=20 \sqrt{5}$ $\mathrm{KE}=\frac{1}{2} \mathrm{mu}^{2}=\frac{1}{2}(0.05)(20 \sqrt{5})^{2}$ $=\frac{1}{2} \times(0.05)(2000)=50 \mathrm{~J}$ Now, $\mathrm{K} \cdot \mathrm{E}^{\prime} =70 \% \mathrm{~K} \cdot \mathrm{E}$ $\mathrm{K} \cdot \mathrm{E}^{\prime} =0.7 \mathrm{KE}=0.7 \times 50$ $=35 \mathrm{~J}$ Then, $\mathrm{K} . \mathrm{E}^{\prime}=\frac{1}{2} \mathrm{mv}^{2}$ $\mathrm{v}=\sqrt{\frac{35 \times 2}{0.05}}=\sqrt{1400}$ Using the equation of motion, $v^{2}=u^{2}-2 g h$ $1400=2000-2(10) h$ $-600=-20 h$ $h=30 \mathrm{~m}$
Shift-II]
Work, Energy and Power
148953
As shown in the figure below, the force $F$ on a particle varies with position in a certain manner. The kinetic energy of the particle at $\mathbf{x}$ $=0$ is $12 \mathrm{~J}$. What is its kinetic energy at $x=14$ ?
1 $42 \mathrm{~J}$
2 $70 \mathrm{~J}$
3 $40 \mathrm{~J}$
4 $0 \mathrm{~J}$
Explanation:
A Given, Kinetic energy (K.E.) $)_{(\mathrm{x}=0)}=12 \mathrm{~J}$ Work done $(\mathrm{W})=$ Area under $(\mathrm{F}-\mathrm{x})$ graph Work done $(\mathrm{W})=$ Area of $\triangle \mathrm{AOB}+$ Area of $\square \mathrm{BCDE}+$ Area of $\square \mathrm{EFGH}$ $\mathrm{W} =\frac{1}{2} \times 6 \times 10+(-5) \times(10-6)+5 \times(14-10)$ $\mathrm{W} =30-20+20$ $\mathrm{~W} =30 \mathrm{~J}$ We know that Work done $(\mathrm{W})=$ Change in kinetic energy Work done $=(\text { K.E. })_{\mathrm{f}}-(\text { K.E. })_{\mathrm{i}}$ So, $30 \mathrm{~J}=(\mathrm{K} . \mathrm{E} .)_{\mathrm{f}}-12 \mathrm{~J}$ (K.E. $)_{\mathrm{f}}=42 \mathrm{~J}$
Assam CEE-2021
Work, Energy and Power
148954
A particle is moving along the $x$-axis and has potential energy $U=1+12 x+6 x^{2}$. If the particle is now released at $x=-5$, the maximum value of $x$ can be
1 -1
2 0
3 1
4 3
Explanation:
D Given that, Potential energy $(U)=1+12 x+6 x^{2}$ $\because \mathrm{F} =\frac{-\mathrm{dU}}{\mathrm{dx}}$ $=\frac{-\mathrm{d}\left(1+12 \mathrm{x}+6 \mathrm{x}^{2}\right)}{\mathrm{dx}}$ $\mathrm{F} =0-12-12 \mathrm{x}$ At equilibrium position - $F=0$ $-12-12 x=0$ $x=-1$ It will oscillate about $\mathrm{x}=-1$ with amplitude of $4 \mathrm{~m}$ $\therefore$ Maximum value of $x$ will be $3 \mathrm{~m}$
Assam CEE-2021
Work, Energy and Power
148955
A machine gun fires 360 bullets per minute. Each bullet travels with velocity of $500 \mathrm{~m} / \mathrm{s}$. If the power of the machine gun is $4.5 \mathrm{~kW}$ then the mass of each bullet is
1 $2 \mathrm{~g}$
2 $5 \mathrm{~g}$
3 $6 \mathrm{~g}$
4 $10 \mathrm{~g}$
Explanation:
C Given, No. of bullets $(n)=360$, Velocity of each bullet $(\mathrm{v})=500 \mathrm{~m} / \mathrm{s}$ We know, Power $(P)=4500$ watt, $t=1$ minute $=60 \mathrm{~s}$ Kinetic energy of 360 bullets (K.E) $=n \times \frac{1}{2} m v^{2}$ $=360 \times \frac{1}{2} \times \mathrm{m} \times(500)^{2}$ $\mathrm{~K} . \mathrm{E}= \left(45 \times 10^{6} \mathrm{~m}\right) \mathrm{J}$ $\text { Power }(\mathrm{P}) =\frac{\mathrm{K} . \mathrm{E}}{\text { Time }}$ $4500 =\frac{45 \times 10^{6} \times \mathrm{m}}{60}$ $\mathrm{~m} =\frac{4500 \times 60}{45 \times 10^{6}}$ $\mathrm{~m} =0.006 \mathrm{~kg}$ $\mathrm{~m} =6 \mathrm{~g}$
148952
When a ball of mass $50 \mathrm{~g}$ is thrown upwards. it raises to a maximum height of $100 \mathrm{~m}$. At what height its K.E. will be reduced to $70 \%$
1 $30 \mathrm{~m}$
2 $40 \mathrm{~m}$
3 $60 \mathrm{~m}$
4 $70 \mathrm{~m}$
Explanation:
A Given, Mass $(\mathrm{m})=50 \mathrm{~g}=0.05 \mathrm{~kg}$ Height $(\mathrm{h})=100 \mathrm{~m}$ At maximum height $\mathrm{v}^{2}=\mathrm{u}^{2}-2 \mathrm{gh}$ $0=\mathrm{u}^{2}-2(10)(100)$ $\mathrm{u}=\sqrt{2000}=20 \sqrt{5}$ $\mathrm{KE}=\frac{1}{2} \mathrm{mu}^{2}=\frac{1}{2}(0.05)(20 \sqrt{5})^{2}$ $=\frac{1}{2} \times(0.05)(2000)=50 \mathrm{~J}$ Now, $\mathrm{K} \cdot \mathrm{E}^{\prime} =70 \% \mathrm{~K} \cdot \mathrm{E}$ $\mathrm{K} \cdot \mathrm{E}^{\prime} =0.7 \mathrm{KE}=0.7 \times 50$ $=35 \mathrm{~J}$ Then, $\mathrm{K} . \mathrm{E}^{\prime}=\frac{1}{2} \mathrm{mv}^{2}$ $\mathrm{v}=\sqrt{\frac{35 \times 2}{0.05}}=\sqrt{1400}$ Using the equation of motion, $v^{2}=u^{2}-2 g h$ $1400=2000-2(10) h$ $-600=-20 h$ $h=30 \mathrm{~m}$
Shift-II]
Work, Energy and Power
148953
As shown in the figure below, the force $F$ on a particle varies with position in a certain manner. The kinetic energy of the particle at $\mathbf{x}$ $=0$ is $12 \mathrm{~J}$. What is its kinetic energy at $x=14$ ?
1 $42 \mathrm{~J}$
2 $70 \mathrm{~J}$
3 $40 \mathrm{~J}$
4 $0 \mathrm{~J}$
Explanation:
A Given, Kinetic energy (K.E.) $)_{(\mathrm{x}=0)}=12 \mathrm{~J}$ Work done $(\mathrm{W})=$ Area under $(\mathrm{F}-\mathrm{x})$ graph Work done $(\mathrm{W})=$ Area of $\triangle \mathrm{AOB}+$ Area of $\square \mathrm{BCDE}+$ Area of $\square \mathrm{EFGH}$ $\mathrm{W} =\frac{1}{2} \times 6 \times 10+(-5) \times(10-6)+5 \times(14-10)$ $\mathrm{W} =30-20+20$ $\mathrm{~W} =30 \mathrm{~J}$ We know that Work done $(\mathrm{W})=$ Change in kinetic energy Work done $=(\text { K.E. })_{\mathrm{f}}-(\text { K.E. })_{\mathrm{i}}$ So, $30 \mathrm{~J}=(\mathrm{K} . \mathrm{E} .)_{\mathrm{f}}-12 \mathrm{~J}$ (K.E. $)_{\mathrm{f}}=42 \mathrm{~J}$
Assam CEE-2021
Work, Energy and Power
148954
A particle is moving along the $x$-axis and has potential energy $U=1+12 x+6 x^{2}$. If the particle is now released at $x=-5$, the maximum value of $x$ can be
1 -1
2 0
3 1
4 3
Explanation:
D Given that, Potential energy $(U)=1+12 x+6 x^{2}$ $\because \mathrm{F} =\frac{-\mathrm{dU}}{\mathrm{dx}}$ $=\frac{-\mathrm{d}\left(1+12 \mathrm{x}+6 \mathrm{x}^{2}\right)}{\mathrm{dx}}$ $\mathrm{F} =0-12-12 \mathrm{x}$ At equilibrium position - $F=0$ $-12-12 x=0$ $x=-1$ It will oscillate about $\mathrm{x}=-1$ with amplitude of $4 \mathrm{~m}$ $\therefore$ Maximum value of $x$ will be $3 \mathrm{~m}$
Assam CEE-2021
Work, Energy and Power
148955
A machine gun fires 360 bullets per minute. Each bullet travels with velocity of $500 \mathrm{~m} / \mathrm{s}$. If the power of the machine gun is $4.5 \mathrm{~kW}$ then the mass of each bullet is
1 $2 \mathrm{~g}$
2 $5 \mathrm{~g}$
3 $6 \mathrm{~g}$
4 $10 \mathrm{~g}$
Explanation:
C Given, No. of bullets $(n)=360$, Velocity of each bullet $(\mathrm{v})=500 \mathrm{~m} / \mathrm{s}$ We know, Power $(P)=4500$ watt, $t=1$ minute $=60 \mathrm{~s}$ Kinetic energy of 360 bullets (K.E) $=n \times \frac{1}{2} m v^{2}$ $=360 \times \frac{1}{2} \times \mathrm{m} \times(500)^{2}$ $\mathrm{~K} . \mathrm{E}= \left(45 \times 10^{6} \mathrm{~m}\right) \mathrm{J}$ $\text { Power }(\mathrm{P}) =\frac{\mathrm{K} . \mathrm{E}}{\text { Time }}$ $4500 =\frac{45 \times 10^{6} \times \mathrm{m}}{60}$ $\mathrm{~m} =\frac{4500 \times 60}{45 \times 10^{6}}$ $\mathrm{~m} =0.006 \mathrm{~kg}$ $\mathrm{~m} =6 \mathrm{~g}$
148952
When a ball of mass $50 \mathrm{~g}$ is thrown upwards. it raises to a maximum height of $100 \mathrm{~m}$. At what height its K.E. will be reduced to $70 \%$
1 $30 \mathrm{~m}$
2 $40 \mathrm{~m}$
3 $60 \mathrm{~m}$
4 $70 \mathrm{~m}$
Explanation:
A Given, Mass $(\mathrm{m})=50 \mathrm{~g}=0.05 \mathrm{~kg}$ Height $(\mathrm{h})=100 \mathrm{~m}$ At maximum height $\mathrm{v}^{2}=\mathrm{u}^{2}-2 \mathrm{gh}$ $0=\mathrm{u}^{2}-2(10)(100)$ $\mathrm{u}=\sqrt{2000}=20 \sqrt{5}$ $\mathrm{KE}=\frac{1}{2} \mathrm{mu}^{2}=\frac{1}{2}(0.05)(20 \sqrt{5})^{2}$ $=\frac{1}{2} \times(0.05)(2000)=50 \mathrm{~J}$ Now, $\mathrm{K} \cdot \mathrm{E}^{\prime} =70 \% \mathrm{~K} \cdot \mathrm{E}$ $\mathrm{K} \cdot \mathrm{E}^{\prime} =0.7 \mathrm{KE}=0.7 \times 50$ $=35 \mathrm{~J}$ Then, $\mathrm{K} . \mathrm{E}^{\prime}=\frac{1}{2} \mathrm{mv}^{2}$ $\mathrm{v}=\sqrt{\frac{35 \times 2}{0.05}}=\sqrt{1400}$ Using the equation of motion, $v^{2}=u^{2}-2 g h$ $1400=2000-2(10) h$ $-600=-20 h$ $h=30 \mathrm{~m}$
Shift-II]
Work, Energy and Power
148953
As shown in the figure below, the force $F$ on a particle varies with position in a certain manner. The kinetic energy of the particle at $\mathbf{x}$ $=0$ is $12 \mathrm{~J}$. What is its kinetic energy at $x=14$ ?
1 $42 \mathrm{~J}$
2 $70 \mathrm{~J}$
3 $40 \mathrm{~J}$
4 $0 \mathrm{~J}$
Explanation:
A Given, Kinetic energy (K.E.) $)_{(\mathrm{x}=0)}=12 \mathrm{~J}$ Work done $(\mathrm{W})=$ Area under $(\mathrm{F}-\mathrm{x})$ graph Work done $(\mathrm{W})=$ Area of $\triangle \mathrm{AOB}+$ Area of $\square \mathrm{BCDE}+$ Area of $\square \mathrm{EFGH}$ $\mathrm{W} =\frac{1}{2} \times 6 \times 10+(-5) \times(10-6)+5 \times(14-10)$ $\mathrm{W} =30-20+20$ $\mathrm{~W} =30 \mathrm{~J}$ We know that Work done $(\mathrm{W})=$ Change in kinetic energy Work done $=(\text { K.E. })_{\mathrm{f}}-(\text { K.E. })_{\mathrm{i}}$ So, $30 \mathrm{~J}=(\mathrm{K} . \mathrm{E} .)_{\mathrm{f}}-12 \mathrm{~J}$ (K.E. $)_{\mathrm{f}}=42 \mathrm{~J}$
Assam CEE-2021
Work, Energy and Power
148954
A particle is moving along the $x$-axis and has potential energy $U=1+12 x+6 x^{2}$. If the particle is now released at $x=-5$, the maximum value of $x$ can be
1 -1
2 0
3 1
4 3
Explanation:
D Given that, Potential energy $(U)=1+12 x+6 x^{2}$ $\because \mathrm{F} =\frac{-\mathrm{dU}}{\mathrm{dx}}$ $=\frac{-\mathrm{d}\left(1+12 \mathrm{x}+6 \mathrm{x}^{2}\right)}{\mathrm{dx}}$ $\mathrm{F} =0-12-12 \mathrm{x}$ At equilibrium position - $F=0$ $-12-12 x=0$ $x=-1$ It will oscillate about $\mathrm{x}=-1$ with amplitude of $4 \mathrm{~m}$ $\therefore$ Maximum value of $x$ will be $3 \mathrm{~m}$
Assam CEE-2021
Work, Energy and Power
148955
A machine gun fires 360 bullets per minute. Each bullet travels with velocity of $500 \mathrm{~m} / \mathrm{s}$. If the power of the machine gun is $4.5 \mathrm{~kW}$ then the mass of each bullet is
1 $2 \mathrm{~g}$
2 $5 \mathrm{~g}$
3 $6 \mathrm{~g}$
4 $10 \mathrm{~g}$
Explanation:
C Given, No. of bullets $(n)=360$, Velocity of each bullet $(\mathrm{v})=500 \mathrm{~m} / \mathrm{s}$ We know, Power $(P)=4500$ watt, $t=1$ minute $=60 \mathrm{~s}$ Kinetic energy of 360 bullets (K.E) $=n \times \frac{1}{2} m v^{2}$ $=360 \times \frac{1}{2} \times \mathrm{m} \times(500)^{2}$ $\mathrm{~K} . \mathrm{E}= \left(45 \times 10^{6} \mathrm{~m}\right) \mathrm{J}$ $\text { Power }(\mathrm{P}) =\frac{\mathrm{K} . \mathrm{E}}{\text { Time }}$ $4500 =\frac{45 \times 10^{6} \times \mathrm{m}}{60}$ $\mathrm{~m} =\frac{4500 \times 60}{45 \times 10^{6}}$ $\mathrm{~m} =0.006 \mathrm{~kg}$ $\mathrm{~m} =6 \mathrm{~g}$
