NEET Test Series from KOTA - 10 Papers In MS WORD
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Work, Energy and Power
149149
A ball of mass $m$ is thrown upward with a velocity v. The air exerts an average resisting force $F$. The velocity with which the ball returns to the thrower is
1 $v \sqrt{\frac{m g}{m g+F}}$
2 $v \sqrt{\frac{F}{m g+F}}$
3 $v \sqrt{\frac{F-m g}{F+m g}}$
4 $v \sqrt{\frac{m g+F}{m g}}$
Explanation:
C At maximum height speed of the ball is 0 By $\frac{1}{2} \mathrm{mv}^{2}=\text { Work Done }$ $\frac{1}{2} \mathrm{mv}^{2}=-\mathrm{FH}-\mathrm{mgH}$ and, Work - Energy theorem, $\frac{1}{2} \mathrm{mv}_{1}^{2}=-\mathrm{FH}+\mathrm{mgH}-$ Divide (1) by (2) $\frac{\frac{1}{2} \mathrm{mv}^{2}}{\frac{1}{2} \mathrm{mv}_{1}^{2}}=\frac{-(\mathrm{FH}+\mathrm{mgH})}{-(\mathrm{FH}-\mathrm{mgH})}$ $\frac{\mathrm{v}^{2}}{\mathrm{v}_{1}^{2}}=\frac{\mathrm{F}+\mathrm{mg}}{\mathrm{F}-\mathrm{mg}}$ $\mathrm{v}_{1}^{2}=\frac{\mathrm{F}-\mathrm{mg}}{\mathrm{F}+\mathrm{mg}} \times \mathrm{v}^{2}$ $\mathrm{v}_{1}=\mathrm{v} \sqrt{\frac{\mathrm{F}-\mathrm{mg}}{\mathrm{F}+\mathrm{mg}}}$
Assam CEE-2016
Work, Energy and Power
149121
Which of the following graphs represent kinetic energy (K), potential energy (U) and height (h) from the ground for a freely falling body?
1
2
3
4
Explanation:
A We know if body is falling freely then total mechanical energy is constant. According to law of conservation of energy $\text { P.E. }+ \text { K.E. }=\text { Constant }$ Potential energy increases and kinetic energy decreases when the height of the particle increases. So, Graph (A) represents kinetic energy (K), potential energy (U) and height (h) from the ground for a freely falling body.
CG PET- 2017
Work, Energy and Power
149123
The work-energy theorem states that the change in
1 Kinetic energy of a particle is equal to the work done on it by the net force
2 Kinetic energy of a particle is equal to the work done by one of forces acting on it
3 Potential energy of a particle is equal to the work done on it by the net force
4 Potential energy of a particle is equal to the work done by one of the forces acting on it
5 Total energy of a particle is equal to the work done on it by the net force
Explanation:
A The work-energy theorem states that the net work done by the forces on an object is equal to the change in its kinetic energy. Thus, $\quad \mathrm{W}=\Delta$ K.E. $=\frac{1}{2} \mathrm{mv}^{2}$
Kerala CEE - 2016
Work, Energy and Power
268594
A lorry and a car moving with the same \(K E\) are brought to rest by applying the same retarding force. Then
149149
A ball of mass $m$ is thrown upward with a velocity v. The air exerts an average resisting force $F$. The velocity with which the ball returns to the thrower is
1 $v \sqrt{\frac{m g}{m g+F}}$
2 $v \sqrt{\frac{F}{m g+F}}$
3 $v \sqrt{\frac{F-m g}{F+m g}}$
4 $v \sqrt{\frac{m g+F}{m g}}$
Explanation:
C At maximum height speed of the ball is 0 By $\frac{1}{2} \mathrm{mv}^{2}=\text { Work Done }$ $\frac{1}{2} \mathrm{mv}^{2}=-\mathrm{FH}-\mathrm{mgH}$ and, Work - Energy theorem, $\frac{1}{2} \mathrm{mv}_{1}^{2}=-\mathrm{FH}+\mathrm{mgH}-$ Divide (1) by (2) $\frac{\frac{1}{2} \mathrm{mv}^{2}}{\frac{1}{2} \mathrm{mv}_{1}^{2}}=\frac{-(\mathrm{FH}+\mathrm{mgH})}{-(\mathrm{FH}-\mathrm{mgH})}$ $\frac{\mathrm{v}^{2}}{\mathrm{v}_{1}^{2}}=\frac{\mathrm{F}+\mathrm{mg}}{\mathrm{F}-\mathrm{mg}}$ $\mathrm{v}_{1}^{2}=\frac{\mathrm{F}-\mathrm{mg}}{\mathrm{F}+\mathrm{mg}} \times \mathrm{v}^{2}$ $\mathrm{v}_{1}=\mathrm{v} \sqrt{\frac{\mathrm{F}-\mathrm{mg}}{\mathrm{F}+\mathrm{mg}}}$
Assam CEE-2016
Work, Energy and Power
149121
Which of the following graphs represent kinetic energy (K), potential energy (U) and height (h) from the ground for a freely falling body?
1
2
3
4
Explanation:
A We know if body is falling freely then total mechanical energy is constant. According to law of conservation of energy $\text { P.E. }+ \text { K.E. }=\text { Constant }$ Potential energy increases and kinetic energy decreases when the height of the particle increases. So, Graph (A) represents kinetic energy (K), potential energy (U) and height (h) from the ground for a freely falling body.
CG PET- 2017
Work, Energy and Power
149123
The work-energy theorem states that the change in
1 Kinetic energy of a particle is equal to the work done on it by the net force
2 Kinetic energy of a particle is equal to the work done by one of forces acting on it
3 Potential energy of a particle is equal to the work done on it by the net force
4 Potential energy of a particle is equal to the work done by one of the forces acting on it
5 Total energy of a particle is equal to the work done on it by the net force
Explanation:
A The work-energy theorem states that the net work done by the forces on an object is equal to the change in its kinetic energy. Thus, $\quad \mathrm{W}=\Delta$ K.E. $=\frac{1}{2} \mathrm{mv}^{2}$
Kerala CEE - 2016
Work, Energy and Power
268594
A lorry and a car moving with the same \(K E\) are brought to rest by applying the same retarding force. Then
149149
A ball of mass $m$ is thrown upward with a velocity v. The air exerts an average resisting force $F$. The velocity with which the ball returns to the thrower is
1 $v \sqrt{\frac{m g}{m g+F}}$
2 $v \sqrt{\frac{F}{m g+F}}$
3 $v \sqrt{\frac{F-m g}{F+m g}}$
4 $v \sqrt{\frac{m g+F}{m g}}$
Explanation:
C At maximum height speed of the ball is 0 By $\frac{1}{2} \mathrm{mv}^{2}=\text { Work Done }$ $\frac{1}{2} \mathrm{mv}^{2}=-\mathrm{FH}-\mathrm{mgH}$ and, Work - Energy theorem, $\frac{1}{2} \mathrm{mv}_{1}^{2}=-\mathrm{FH}+\mathrm{mgH}-$ Divide (1) by (2) $\frac{\frac{1}{2} \mathrm{mv}^{2}}{\frac{1}{2} \mathrm{mv}_{1}^{2}}=\frac{-(\mathrm{FH}+\mathrm{mgH})}{-(\mathrm{FH}-\mathrm{mgH})}$ $\frac{\mathrm{v}^{2}}{\mathrm{v}_{1}^{2}}=\frac{\mathrm{F}+\mathrm{mg}}{\mathrm{F}-\mathrm{mg}}$ $\mathrm{v}_{1}^{2}=\frac{\mathrm{F}-\mathrm{mg}}{\mathrm{F}+\mathrm{mg}} \times \mathrm{v}^{2}$ $\mathrm{v}_{1}=\mathrm{v} \sqrt{\frac{\mathrm{F}-\mathrm{mg}}{\mathrm{F}+\mathrm{mg}}}$
Assam CEE-2016
Work, Energy and Power
149121
Which of the following graphs represent kinetic energy (K), potential energy (U) and height (h) from the ground for a freely falling body?
1
2
3
4
Explanation:
A We know if body is falling freely then total mechanical energy is constant. According to law of conservation of energy $\text { P.E. }+ \text { K.E. }=\text { Constant }$ Potential energy increases and kinetic energy decreases when the height of the particle increases. So, Graph (A) represents kinetic energy (K), potential energy (U) and height (h) from the ground for a freely falling body.
CG PET- 2017
Work, Energy and Power
149123
The work-energy theorem states that the change in
1 Kinetic energy of a particle is equal to the work done on it by the net force
2 Kinetic energy of a particle is equal to the work done by one of forces acting on it
3 Potential energy of a particle is equal to the work done on it by the net force
4 Potential energy of a particle is equal to the work done by one of the forces acting on it
5 Total energy of a particle is equal to the work done on it by the net force
Explanation:
A The work-energy theorem states that the net work done by the forces on an object is equal to the change in its kinetic energy. Thus, $\quad \mathrm{W}=\Delta$ K.E. $=\frac{1}{2} \mathrm{mv}^{2}$
Kerala CEE - 2016
Work, Energy and Power
268594
A lorry and a car moving with the same \(K E\) are brought to rest by applying the same retarding force. Then
149149
A ball of mass $m$ is thrown upward with a velocity v. The air exerts an average resisting force $F$. The velocity with which the ball returns to the thrower is
1 $v \sqrt{\frac{m g}{m g+F}}$
2 $v \sqrt{\frac{F}{m g+F}}$
3 $v \sqrt{\frac{F-m g}{F+m g}}$
4 $v \sqrt{\frac{m g+F}{m g}}$
Explanation:
C At maximum height speed of the ball is 0 By $\frac{1}{2} \mathrm{mv}^{2}=\text { Work Done }$ $\frac{1}{2} \mathrm{mv}^{2}=-\mathrm{FH}-\mathrm{mgH}$ and, Work - Energy theorem, $\frac{1}{2} \mathrm{mv}_{1}^{2}=-\mathrm{FH}+\mathrm{mgH}-$ Divide (1) by (2) $\frac{\frac{1}{2} \mathrm{mv}^{2}}{\frac{1}{2} \mathrm{mv}_{1}^{2}}=\frac{-(\mathrm{FH}+\mathrm{mgH})}{-(\mathrm{FH}-\mathrm{mgH})}$ $\frac{\mathrm{v}^{2}}{\mathrm{v}_{1}^{2}}=\frac{\mathrm{F}+\mathrm{mg}}{\mathrm{F}-\mathrm{mg}}$ $\mathrm{v}_{1}^{2}=\frac{\mathrm{F}-\mathrm{mg}}{\mathrm{F}+\mathrm{mg}} \times \mathrm{v}^{2}$ $\mathrm{v}_{1}=\mathrm{v} \sqrt{\frac{\mathrm{F}-\mathrm{mg}}{\mathrm{F}+\mathrm{mg}}}$
Assam CEE-2016
Work, Energy and Power
149121
Which of the following graphs represent kinetic energy (K), potential energy (U) and height (h) from the ground for a freely falling body?
1
2
3
4
Explanation:
A We know if body is falling freely then total mechanical energy is constant. According to law of conservation of energy $\text { P.E. }+ \text { K.E. }=\text { Constant }$ Potential energy increases and kinetic energy decreases when the height of the particle increases. So, Graph (A) represents kinetic energy (K), potential energy (U) and height (h) from the ground for a freely falling body.
CG PET- 2017
Work, Energy and Power
149123
The work-energy theorem states that the change in
1 Kinetic energy of a particle is equal to the work done on it by the net force
2 Kinetic energy of a particle is equal to the work done by one of forces acting on it
3 Potential energy of a particle is equal to the work done on it by the net force
4 Potential energy of a particle is equal to the work done by one of the forces acting on it
5 Total energy of a particle is equal to the work done on it by the net force
Explanation:
A The work-energy theorem states that the net work done by the forces on an object is equal to the change in its kinetic energy. Thus, $\quad \mathrm{W}=\Delta$ K.E. $=\frac{1}{2} \mathrm{mv}^{2}$
Kerala CEE - 2016
Work, Energy and Power
268594
A lorry and a car moving with the same \(K E\) are brought to rest by applying the same retarding force. Then
