QBManager

Work, Energy and Power

149103 A $750 kg$ boat is $10 m$ long and is floating without motion on still water. A man of mass $80 kg$ is at one end and if he runs to another end of the boat and stops, the displacement of boat is

1

1.8 m

in the direction of displacement of man

2

0.96 m

in the direction of opposite to the displacement of man

3

0.96 m

in the direction of displacement of the man

4

1.8 m

in the direction opposite to displacement of man

Explanation:

B Given,
Mass of $man (m_{1}) = 80 kg$
Mass of boat $(m_{2}) = 750 kg$
Length of boat $(l_{b}) = 10 m$

Displacement of man w.r.t. boat,
$x = l_{b} (\frac{m_{1}}{m_{1} + m_{2}}) = - 10 \times (\frac{80}{80 + 750})$
$= - 10 \times (\frac{80}{830})$
$= - 0.96 m$
Therefore, the displacement of the boat is $0.96 m$ in the direction of opposite to the displacement of man.

TS EAMCET 06.08.2021

Work, Energy and Power

149104 A car driver is trying to jump across a path as shown in figure by driving horizontally off a cliff ' $X$ ' at the speed $10 m / s$ . When he touches peak $Z$ (ignore air resistance), what would be speed? $($ use $g = 10 m / s^{2})$

1

30 m / s

2

40 m / s

3

15 m / s

4

50 m / s

Explanation:

A Here all forces are of conservative nature as we can neglect air resistance and no external force is acting on this system.
From energy conservation law,
If all forces are of conservative nature and no external force is acting on system, then mechanical energy at $x$ and $z$ will be equal-
$(M.E.)_{x} = (M.E.)_{z}$
${mgh}_{x} + \frac{1}{2} {mv}_{x}^{2} = {mgh}_{z} + \frac{1}{2} {mv}_{z}^{2}$
$mg (100) + \frac{1}{2} m (10)^{2} = mg (60) + \frac{1}{2} {mv}_{z}^{2}$
$100 g + \frac{1}{2} \times 100 = 60 g + \frac{1}{2} v_{z}^{2} (g = 10 m / s^{2})$
$1000 + 50 = 600 + \frac{1}{2} v_{z}^{2}$
$2 (1050 - 600) = v_{z}^{2}$
$900 = v_{z}^{2}$
$v_{z} = 30 m / s$

TS EAMCET 04.08.2021

Work, Energy and Power

149105 The Work-Energy theorem

1 Does not hold in all inertial frames

2 Is independent of Newton's second law

3 May be viewed as a scalar form of Newton's second law

4 Cannot be extended to non-inertial frames

5 In independent of Newton's third law

Explanation:

C The Work-Energy theorem
The total work done by all the forces combined-done an object is equal to the change in the kinetic energy of the object. After the net force is removed (no more work is being done) the object's total energy is altered as a result of the work that was done. May be viewed as a scalar form of Newton's second law.
Newton's second law state that,
$F = ma = \frac{dB}{dt}$
The work energy theorem statement,
$work done = F.ds = Δ K.E. = \frac{1}{2} m (v_{f}^{2} - v_{i}^{2})$

Kerala CEE 2020

Work, Energy and Power

149106 Which of the following statements is false?

1 Momentum is conserved in all types of collisions

2 Energy is conserved in all types of collisions

3 During elastic collisions conservative forces are involved

4 Work-energy theorem is not applicable to inelastic collisions

Explanation:

D Momentum of all types of collisions are conserved $m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$
- Energy of all types of collision are conserved
$\frac{1}{2} m_{1} u_{1}^{2} + \frac{1}{2} m_{2} u_{2}^{2} = \frac{1}{2} m_{1} v_{1}^{2} + \frac{1}{2} m_{2} v_{2}^{2}$
- During elastic collisions conservative forces are involved.
- Work energy theorem is applicable for all type of collision.
Hence, option (d) is false.

AP EAMCET-24.09.2020

Work, Energy and Power

149103 A $750 kg$ boat is $10 m$ long and is floating without motion on still water. A man of mass $80 kg$ is at one end and if he runs to another end of the boat and stops, the displacement of boat is

1

1.8 m

in the direction of displacement of man

2

0.96 m

in the direction of opposite to the displacement of man

3

0.96 m

in the direction of displacement of the man

4

1.8 m

in the direction opposite to displacement of man

Explanation:

B Given,
Mass of $man (m_{1}) = 80 kg$
Mass of boat $(m_{2}) = 750 kg$
Length of boat $(l_{b}) = 10 m$

Displacement of man w.r.t. boat,
$x = l_{b} (\frac{m_{1}}{m_{1} + m_{2}}) = - 10 \times (\frac{80}{80 + 750})$
$= - 10 \times (\frac{80}{830})$
$= - 0.96 m$
Therefore, the displacement of the boat is $0.96 m$ in the direction of opposite to the displacement of man.

TS EAMCET 06.08.2021

Work, Energy and Power

149104 A car driver is trying to jump across a path as shown in figure by driving horizontally off a cliff ' $X$ ' at the speed $10 m / s$ . When he touches peak $Z$ (ignore air resistance), what would be speed? $($ use $g = 10 m / s^{2})$

1

30 m / s

2

40 m / s

3

15 m / s

4

50 m / s

Explanation:

A Here all forces are of conservative nature as we can neglect air resistance and no external force is acting on this system.
From energy conservation law,
If all forces are of conservative nature and no external force is acting on system, then mechanical energy at $x$ and $z$ will be equal-
$(M.E.)_{x} = (M.E.)_{z}$
${mgh}_{x} + \frac{1}{2} {mv}_{x}^{2} = {mgh}_{z} + \frac{1}{2} {mv}_{z}^{2}$
$mg (100) + \frac{1}{2} m (10)^{2} = mg (60) + \frac{1}{2} {mv}_{z}^{2}$
$100 g + \frac{1}{2} \times 100 = 60 g + \frac{1}{2} v_{z}^{2} (g = 10 m / s^{2})$
$1000 + 50 = 600 + \frac{1}{2} v_{z}^{2}$
$2 (1050 - 600) = v_{z}^{2}$
$900 = v_{z}^{2}$
$v_{z} = 30 m / s$

TS EAMCET 04.08.2021

Work, Energy and Power

149105 The Work-Energy theorem

1 Does not hold in all inertial frames

2 Is independent of Newton's second law

3 May be viewed as a scalar form of Newton's second law

4 Cannot be extended to non-inertial frames

5 In independent of Newton's third law

Explanation:

C The Work-Energy theorem
The total work done by all the forces combined-done an object is equal to the change in the kinetic energy of the object. After the net force is removed (no more work is being done) the object's total energy is altered as a result of the work that was done. May be viewed as a scalar form of Newton's second law.
Newton's second law state that,
$F = ma = \frac{dB}{dt}$
The work energy theorem statement,
$work done = F.ds = Δ K.E. = \frac{1}{2} m (v_{f}^{2} - v_{i}^{2})$

Kerala CEE 2020

Work, Energy and Power

149106 Which of the following statements is false?

1 Momentum is conserved in all types of collisions

2 Energy is conserved in all types of collisions

3 During elastic collisions conservative forces are involved

4 Work-energy theorem is not applicable to inelastic collisions

Explanation:

D Momentum of all types of collisions are conserved $m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$
- Energy of all types of collision are conserved
$\frac{1}{2} m_{1} u_{1}^{2} + \frac{1}{2} m_{2} u_{2}^{2} = \frac{1}{2} m_{1} v_{1}^{2} + \frac{1}{2} m_{2} v_{2}^{2}$
- During elastic collisions conservative forces are involved.
- Work energy theorem is applicable for all type of collision.
Hence, option (d) is false.

AP EAMCET-24.09.2020

Work, Energy and Power

149103 A $750 kg$ boat is $10 m$ long and is floating without motion on still water. A man of mass $80 kg$ is at one end and if he runs to another end of the boat and stops, the displacement of boat is

1

1.8 m

in the direction of displacement of man

2

0.96 m

in the direction of opposite to the displacement of man

3

0.96 m

in the direction of displacement of the man

4

1.8 m

in the direction opposite to displacement of man

Explanation:

B Given,
Mass of $man (m_{1}) = 80 kg$
Mass of boat $(m_{2}) = 750 kg$
Length of boat $(l_{b}) = 10 m$

Displacement of man w.r.t. boat,
$x = l_{b} (\frac{m_{1}}{m_{1} + m_{2}}) = - 10 \times (\frac{80}{80 + 750})$
$= - 10 \times (\frac{80}{830})$
$= - 0.96 m$
Therefore, the displacement of the boat is $0.96 m$ in the direction of opposite to the displacement of man.

TS EAMCET 06.08.2021

Work, Energy and Power

149104 A car driver is trying to jump across a path as shown in figure by driving horizontally off a cliff ' $X$ ' at the speed $10 m / s$ . When he touches peak $Z$ (ignore air resistance), what would be speed? $($ use $g = 10 m / s^{2})$

1

30 m / s

2

40 m / s

3

15 m / s

4

50 m / s

Explanation:

A Here all forces are of conservative nature as we can neglect air resistance and no external force is acting on this system.
From energy conservation law,
If all forces are of conservative nature and no external force is acting on system, then mechanical energy at $x$ and $z$ will be equal-
$(M.E.)_{x} = (M.E.)_{z}$
${mgh}_{x} + \frac{1}{2} {mv}_{x}^{2} = {mgh}_{z} + \frac{1}{2} {mv}_{z}^{2}$
$mg (100) + \frac{1}{2} m (10)^{2} = mg (60) + \frac{1}{2} {mv}_{z}^{2}$
$100 g + \frac{1}{2} \times 100 = 60 g + \frac{1}{2} v_{z}^{2} (g = 10 m / s^{2})$
$1000 + 50 = 600 + \frac{1}{2} v_{z}^{2}$
$2 (1050 - 600) = v_{z}^{2}$
$900 = v_{z}^{2}$
$v_{z} = 30 m / s$

TS EAMCET 04.08.2021

Work, Energy and Power

149105 The Work-Energy theorem

1 Does not hold in all inertial frames

2 Is independent of Newton's second law

3 May be viewed as a scalar form of Newton's second law

4 Cannot be extended to non-inertial frames

5 In independent of Newton's third law

Explanation:

C The Work-Energy theorem
The total work done by all the forces combined-done an object is equal to the change in the kinetic energy of the object. After the net force is removed (no more work is being done) the object's total energy is altered as a result of the work that was done. May be viewed as a scalar form of Newton's second law.
Newton's second law state that,
$F = ma = \frac{dB}{dt}$
The work energy theorem statement,
$work done = F.ds = Δ K.E. = \frac{1}{2} m (v_{f}^{2} - v_{i}^{2})$

Kerala CEE 2020

Work, Energy and Power

149106 Which of the following statements is false?

1 Momentum is conserved in all types of collisions

2 Energy is conserved in all types of collisions

3 During elastic collisions conservative forces are involved

4 Work-energy theorem is not applicable to inelastic collisions

Explanation:

D Momentum of all types of collisions are conserved $m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$
- Energy of all types of collision are conserved
$\frac{1}{2} m_{1} u_{1}^{2} + \frac{1}{2} m_{2} u_{2}^{2} = \frac{1}{2} m_{1} v_{1}^{2} + \frac{1}{2} m_{2} v_{2}^{2}$
- During elastic collisions conservative forces are involved.
- Work energy theorem is applicable for all type of collision.
Hence, option (d) is false.

AP EAMCET-24.09.2020

Work, Energy and Power

149103 A $750 kg$ boat is $10 m$ long and is floating without motion on still water. A man of mass $80 kg$ is at one end and if he runs to another end of the boat and stops, the displacement of boat is

1

1.8 m

in the direction of displacement of man

2

0.96 m

in the direction of opposite to the displacement of man

3

0.96 m

in the direction of displacement of the man

4

1.8 m

in the direction opposite to displacement of man

Explanation:

B Given,
Mass of $man (m_{1}) = 80 kg$
Mass of boat $(m_{2}) = 750 kg$
Length of boat $(l_{b}) = 10 m$

Displacement of man w.r.t. boat,
$x = l_{b} (\frac{m_{1}}{m_{1} + m_{2}}) = - 10 \times (\frac{80}{80 + 750})$
$= - 10 \times (\frac{80}{830})$
$= - 0.96 m$
Therefore, the displacement of the boat is $0.96 m$ in the direction of opposite to the displacement of man.

TS EAMCET 06.08.2021

Work, Energy and Power

149104 A car driver is trying to jump across a path as shown in figure by driving horizontally off a cliff ' $X$ ' at the speed $10 m / s$ . When he touches peak $Z$ (ignore air resistance), what would be speed? $($ use $g = 10 m / s^{2})$

1

30 m / s

2

40 m / s

3

15 m / s

4

50 m / s

Explanation:

A Here all forces are of conservative nature as we can neglect air resistance and no external force is acting on this system.
From energy conservation law,
If all forces are of conservative nature and no external force is acting on system, then mechanical energy at $x$ and $z$ will be equal-
$(M.E.)_{x} = (M.E.)_{z}$
${mgh}_{x} + \frac{1}{2} {mv}_{x}^{2} = {mgh}_{z} + \frac{1}{2} {mv}_{z}^{2}$
$mg (100) + \frac{1}{2} m (10)^{2} = mg (60) + \frac{1}{2} {mv}_{z}^{2}$
$100 g + \frac{1}{2} \times 100 = 60 g + \frac{1}{2} v_{z}^{2} (g = 10 m / s^{2})$
$1000 + 50 = 600 + \frac{1}{2} v_{z}^{2}$
$2 (1050 - 600) = v_{z}^{2}$
$900 = v_{z}^{2}$
$v_{z} = 30 m / s$

TS EAMCET 04.08.2021

Work, Energy and Power

149105 The Work-Energy theorem

1 Does not hold in all inertial frames

2 Is independent of Newton's second law

3 May be viewed as a scalar form of Newton's second law

4 Cannot be extended to non-inertial frames

5 In independent of Newton's third law

Explanation:

C The Work-Energy theorem
The total work done by all the forces combined-done an object is equal to the change in the kinetic energy of the object. After the net force is removed (no more work is being done) the object's total energy is altered as a result of the work that was done. May be viewed as a scalar form of Newton's second law.
Newton's second law state that,
$F = ma = \frac{dB}{dt}$
The work energy theorem statement,
$work done = F.ds = Δ K.E. = \frac{1}{2} m (v_{f}^{2} - v_{i}^{2})$

Kerala CEE 2020

Work, Energy and Power

149106 Which of the following statements is false?

1 Momentum is conserved in all types of collisions

2 Energy is conserved in all types of collisions

3 During elastic collisions conservative forces are involved

4 Work-energy theorem is not applicable to inelastic collisions

Explanation:

D Momentum of all types of collisions are conserved $m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$
- Energy of all types of collision are conserved
$\frac{1}{2} m_{1} u_{1}^{2} + \frac{1}{2} m_{2} u_{2}^{2} = \frac{1}{2} m_{1} v_{1}^{2} + \frac{1}{2} m_{2} v_{2}^{2}$
- During elastic collisions conservative forces are involved.
- Work energy theorem is applicable for all type of collision.
Hence, option (d) is false.

AP EAMCET-24.09.2020