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PHXI06:WORK ENERGY AND POWER

355634 A block of mass $5 k g$ slides down a rough inclined surface. The angle of inclination is $45^{\circ}$ . The coefficient of sliding friction is 0.20 . When the block slides $10 m$ , the work done on the block by force of friction is:

1

50 \sqrt{2} J

2

- 50 \sqrt{2} J

3

50 J

4

- 50 J

Explanation:

supporting img
$W_{F} = μ N x$
$W_{F} = μ m g \cos θ x$
$= 0.2 \times m g \cos θ \times 10$
$= - 0.2 \times 5 \times 10 \times \frac{1}{\sqrt{2}} \times 10 \cos 180^{\circ}$
$= 50 \sqrt{2} \cos 180^{\circ}$
$W_{F} = - 50 \sqrt{2}$ .

PHXI06:WORK ENERGY AND POWER

355635 A 1 $k g$ mass is projected down a rough circular track placed in vertical plane as shown. The speed of the mass at point $A$ is 3 $m / s$ and at point $B$ , it is 60 $m / s$ . How much work is done on the mass between $A$ and $B$ by the force of friction?
supporting img

1

- 8.5 J

2

- 7.5 J

3

- 24 J

4

- 6.5 J

Explanation:

The $K . E$ at $A$ and $B$ are
$\begin{aligned} K_{A} = \frac{1}{2} \times 1 \times 3^{2} = 4 \cdot 5 J \\ K_{B} = \frac{1}{2} \times 1 \times 6^{2} = 18 J \\ K_{B} - K_{A} = 13 \cdot 5 J \\ Δ U_{B A} = m g h = 1 \times 10 \times 2 = 20 \\ | W_{f} | = 20 - 13 \cdot 5 = 6.5 J \end{aligned}$

PHXI06:WORK ENERGY AND POWER

355636 A particle experiences a variable force $\vec{F} = (4 x \hat{i} + 3 y^{2} \hat{j})$ in a horizontal $x$ - $y$ plane. Assume distance in meter and force in newton. If the particle moves from point $(1, 2)$ to point $(2, 3)$ in the $x - y$ plane, the Kinetic energy changes by

1 50.0

J

2 12.5

J

3 25.0

J

4 0

J

Explanation:

$F = 4 x \hat{i} + 3 y^{2} \hat{j}$
We use work-energy theorem
Work done $=$ change in kinetic energy
$W = Δ E_{K}$
Work $W = \int \vec{F} \cdot (d x \hat{i} + d y \hat{j})$
$\begin{aligned} = \int_{1}^{2} 4 x d x + \int_{2}^{3} 3 y^{2} d x \\ \Rightarrow W = {(2 x^{2})}_{1}^{2} + {(y^{3})}_{2}^{3} \\ = (8 - 2) + (27 - 8) \\ = 6 + 19 = 25 J \\ \Rightarrow Δ E_{K} = 25 J \end{aligned}$

JEE - 2022

PHXI06:WORK ENERGY AND POWER

355637 Two cylindrical vessel of equal cross-sectional area contain water upto height $h_{1}$ and $h_{2}$ . The vessels are interconnected so that the levels in them become equal. The work done by the forces of gravity during the process is:

1

ρ A {(\frac{h_{1} + h_{2}}{2})}^{2} g

2

\frac{ρ A h_{1} h_{2}}{2} g

3 Zero

4

ρ A {(\frac{h_{1} - h_{2}}{2})}^{2} g

Explanation:

When level becomes equal then
$h = \frac{h_{1} + h_{2}}{2}$
Work done by gravity $=$ change in potential energy of the water column
$\begin{aligned} = 2 (A h ρ) g \frac{h}{2} - [A h_{1} ρ g \frac{h_{1}}{2} + A h_{2} ρ \frac{g h_{2}}{2}] \\ = A ρ g h^{2} - [A ρ g \frac{h_{1}^{2}}{2} + \frac{A ρ g h_{2}^{2}}{2}] \\ = A ρ g [{(\frac{h_{1} + h_{2}}{2})}^{2} - \frac{h_{1}^{2} + h_{2}^{2}}{2}] \\ = A ρ g {[\frac{h_{1} - h_{2}}{2}]}^{2} \end{aligned}$

PHXI06:WORK ENERGY AND POWER

355634 A block of mass $5 k g$ slides down a rough inclined surface. The angle of inclination is $45^{\circ}$ . The coefficient of sliding friction is 0.20 . When the block slides $10 m$ , the work done on the block by force of friction is:

1

50 \sqrt{2} J

2

- 50 \sqrt{2} J

3

50 J

4

- 50 J

Explanation:

supporting img
$W_{F} = μ N x$
$W_{F} = μ m g \cos θ x$
$= 0.2 \times m g \cos θ \times 10$
$= - 0.2 \times 5 \times 10 \times \frac{1}{\sqrt{2}} \times 10 \cos 180^{\circ}$
$= 50 \sqrt{2} \cos 180^{\circ}$
$W_{F} = - 50 \sqrt{2}$ .

PHXI06:WORK ENERGY AND POWER

355635 A 1 $k g$ mass is projected down a rough circular track placed in vertical plane as shown. The speed of the mass at point $A$ is 3 $m / s$ and at point $B$ , it is 60 $m / s$ . How much work is done on the mass between $A$ and $B$ by the force of friction?
supporting img

1

- 8.5 J

2

- 7.5 J

3

- 24 J

4

- 6.5 J

Explanation:

The $K . E$ at $A$ and $B$ are
$\begin{aligned} K_{A} = \frac{1}{2} \times 1 \times 3^{2} = 4 \cdot 5 J \\ K_{B} = \frac{1}{2} \times 1 \times 6^{2} = 18 J \\ K_{B} - K_{A} = 13 \cdot 5 J \\ Δ U_{B A} = m g h = 1 \times 10 \times 2 = 20 \\ | W_{f} | = 20 - 13 \cdot 5 = 6.5 J \end{aligned}$

PHXI06:WORK ENERGY AND POWER

355636 A particle experiences a variable force $\vec{F} = (4 x \hat{i} + 3 y^{2} \hat{j})$ in a horizontal $x$ - $y$ plane. Assume distance in meter and force in newton. If the particle moves from point $(1, 2)$ to point $(2, 3)$ in the $x - y$ plane, the Kinetic energy changes by

1 50.0

J

2 12.5

J

3 25.0

J

4 0

J

Explanation:

$F = 4 x \hat{i} + 3 y^{2} \hat{j}$
We use work-energy theorem
Work done $=$ change in kinetic energy
$W = Δ E_{K}$
Work $W = \int \vec{F} \cdot (d x \hat{i} + d y \hat{j})$
$\begin{aligned} = \int_{1}^{2} 4 x d x + \int_{2}^{3} 3 y^{2} d x \\ \Rightarrow W = {(2 x^{2})}_{1}^{2} + {(y^{3})}_{2}^{3} \\ = (8 - 2) + (27 - 8) \\ = 6 + 19 = 25 J \\ \Rightarrow Δ E_{K} = 25 J \end{aligned}$

JEE - 2022

PHXI06:WORK ENERGY AND POWER

355637 Two cylindrical vessel of equal cross-sectional area contain water upto height $h_{1}$ and $h_{2}$ . The vessels are interconnected so that the levels in them become equal. The work done by the forces of gravity during the process is:

1

ρ A {(\frac{h_{1} + h_{2}}{2})}^{2} g

2

\frac{ρ A h_{1} h_{2}}{2} g

3 Zero

4

ρ A {(\frac{h_{1} - h_{2}}{2})}^{2} g

Explanation:

When level becomes equal then
$h = \frac{h_{1} + h_{2}}{2}$
Work done by gravity $=$ change in potential energy of the water column
$\begin{aligned} = 2 (A h ρ) g \frac{h}{2} - [A h_{1} ρ g \frac{h_{1}}{2} + A h_{2} ρ \frac{g h_{2}}{2}] \\ = A ρ g h^{2} - [A ρ g \frac{h_{1}^{2}}{2} + \frac{A ρ g h_{2}^{2}}{2}] \\ = A ρ g [{(\frac{h_{1} + h_{2}}{2})}^{2} - \frac{h_{1}^{2} + h_{2}^{2}}{2}] \\ = A ρ g {[\frac{h_{1} - h_{2}}{2}]}^{2} \end{aligned}$

PHXI06:WORK ENERGY AND POWER

355634 A block of mass $5 k g$ slides down a rough inclined surface. The angle of inclination is $45^{\circ}$ . The coefficient of sliding friction is 0.20 . When the block slides $10 m$ , the work done on the block by force of friction is:

1

50 \sqrt{2} J

2

- 50 \sqrt{2} J

3

50 J

4

- 50 J

Explanation:

supporting img
$W_{F} = μ N x$
$W_{F} = μ m g \cos θ x$
$= 0.2 \times m g \cos θ \times 10$
$= - 0.2 \times 5 \times 10 \times \frac{1}{\sqrt{2}} \times 10 \cos 180^{\circ}$
$= 50 \sqrt{2} \cos 180^{\circ}$
$W_{F} = - 50 \sqrt{2}$ .

PHXI06:WORK ENERGY AND POWER

355635 A 1 $k g$ mass is projected down a rough circular track placed in vertical plane as shown. The speed of the mass at point $A$ is 3 $m / s$ and at point $B$ , it is 60 $m / s$ . How much work is done on the mass between $A$ and $B$ by the force of friction?
supporting img

1

- 8.5 J

2

- 7.5 J

3

- 24 J

4

- 6.5 J

Explanation:

The $K . E$ at $A$ and $B$ are
$\begin{aligned} K_{A} = \frac{1}{2} \times 1 \times 3^{2} = 4 \cdot 5 J \\ K_{B} = \frac{1}{2} \times 1 \times 6^{2} = 18 J \\ K_{B} - K_{A} = 13 \cdot 5 J \\ Δ U_{B A} = m g h = 1 \times 10 \times 2 = 20 \\ | W_{f} | = 20 - 13 \cdot 5 = 6.5 J \end{aligned}$

PHXI06:WORK ENERGY AND POWER

355636 A particle experiences a variable force $\vec{F} = (4 x \hat{i} + 3 y^{2} \hat{j})$ in a horizontal $x$ - $y$ plane. Assume distance in meter and force in newton. If the particle moves from point $(1, 2)$ to point $(2, 3)$ in the $x - y$ plane, the Kinetic energy changes by

1 50.0

J

2 12.5

J

3 25.0

J

4 0

J

Explanation:

$F = 4 x \hat{i} + 3 y^{2} \hat{j}$
We use work-energy theorem
Work done $=$ change in kinetic energy
$W = Δ E_{K}$
Work $W = \int \vec{F} \cdot (d x \hat{i} + d y \hat{j})$
$\begin{aligned} = \int_{1}^{2} 4 x d x + \int_{2}^{3} 3 y^{2} d x \\ \Rightarrow W = {(2 x^{2})}_{1}^{2} + {(y^{3})}_{2}^{3} \\ = (8 - 2) + (27 - 8) \\ = 6 + 19 = 25 J \\ \Rightarrow Δ E_{K} = 25 J \end{aligned}$

JEE - 2022

PHXI06:WORK ENERGY AND POWER

355637 Two cylindrical vessel of equal cross-sectional area contain water upto height $h_{1}$ and $h_{2}$ . The vessels are interconnected so that the levels in them become equal. The work done by the forces of gravity during the process is:

1

ρ A {(\frac{h_{1} + h_{2}}{2})}^{2} g

2

\frac{ρ A h_{1} h_{2}}{2} g

3 Zero

4

ρ A {(\frac{h_{1} - h_{2}}{2})}^{2} g

Explanation:

When level becomes equal then
$h = \frac{h_{1} + h_{2}}{2}$
Work done by gravity $=$ change in potential energy of the water column
$\begin{aligned} = 2 (A h ρ) g \frac{h}{2} - [A h_{1} ρ g \frac{h_{1}}{2} + A h_{2} ρ \frac{g h_{2}}{2}] \\ = A ρ g h^{2} - [A ρ g \frac{h_{1}^{2}}{2} + \frac{A ρ g h_{2}^{2}}{2}] \\ = A ρ g [{(\frac{h_{1} + h_{2}}{2})}^{2} - \frac{h_{1}^{2} + h_{2}^{2}}{2}] \\ = A ρ g {[\frac{h_{1} - h_{2}}{2}]}^{2} \end{aligned}$

PHXI06:WORK ENERGY AND POWER

355634 A block of mass $5 k g$ slides down a rough inclined surface. The angle of inclination is $45^{\circ}$ . The coefficient of sliding friction is 0.20 . When the block slides $10 m$ , the work done on the block by force of friction is:

1

50 \sqrt{2} J

2

- 50 \sqrt{2} J

3

50 J

4

- 50 J

Explanation:

supporting img
$W_{F} = μ N x$
$W_{F} = μ m g \cos θ x$
$= 0.2 \times m g \cos θ \times 10$
$= - 0.2 \times 5 \times 10 \times \frac{1}{\sqrt{2}} \times 10 \cos 180^{\circ}$
$= 50 \sqrt{2} \cos 180^{\circ}$
$W_{F} = - 50 \sqrt{2}$ .

PHXI06:WORK ENERGY AND POWER

355635 A 1 $k g$ mass is projected down a rough circular track placed in vertical plane as shown. The speed of the mass at point $A$ is 3 $m / s$ and at point $B$ , it is 60 $m / s$ . How much work is done on the mass between $A$ and $B$ by the force of friction?
supporting img

1

- 8.5 J

2

- 7.5 J

3

- 24 J

4

- 6.5 J

Explanation:

The $K . E$ at $A$ and $B$ are
$\begin{aligned} K_{A} = \frac{1}{2} \times 1 \times 3^{2} = 4 \cdot 5 J \\ K_{B} = \frac{1}{2} \times 1 \times 6^{2} = 18 J \\ K_{B} - K_{A} = 13 \cdot 5 J \\ Δ U_{B A} = m g h = 1 \times 10 \times 2 = 20 \\ | W_{f} | = 20 - 13 \cdot 5 = 6.5 J \end{aligned}$

PHXI06:WORK ENERGY AND POWER

355636 A particle experiences a variable force $\vec{F} = (4 x \hat{i} + 3 y^{2} \hat{j})$ in a horizontal $x$ - $y$ plane. Assume distance in meter and force in newton. If the particle moves from point $(1, 2)$ to point $(2, 3)$ in the $x - y$ plane, the Kinetic energy changes by

1 50.0

J

2 12.5

J

3 25.0

J

4 0

J

Explanation:

$F = 4 x \hat{i} + 3 y^{2} \hat{j}$
We use work-energy theorem
Work done $=$ change in kinetic energy
$W = Δ E_{K}$
Work $W = \int \vec{F} \cdot (d x \hat{i} + d y \hat{j})$
$\begin{aligned} = \int_{1}^{2} 4 x d x + \int_{2}^{3} 3 y^{2} d x \\ \Rightarrow W = {(2 x^{2})}_{1}^{2} + {(y^{3})}_{2}^{3} \\ = (8 - 2) + (27 - 8) \\ = 6 + 19 = 25 J \\ \Rightarrow Δ E_{K} = 25 J \end{aligned}$

JEE - 2022

PHXI06:WORK ENERGY AND POWER

355637 Two cylindrical vessel of equal cross-sectional area contain water upto height $h_{1}$ and $h_{2}$ . The vessels are interconnected so that the levels in them become equal. The work done by the forces of gravity during the process is:

1

ρ A {(\frac{h_{1} + h_{2}}{2})}^{2} g

2

\frac{ρ A h_{1} h_{2}}{2} g

3 Zero

4

ρ A {(\frac{h_{1} - h_{2}}{2})}^{2} g

Explanation:

When level becomes equal then
$h = \frac{h_{1} + h_{2}}{2}$
Work done by gravity $=$ change in potential energy of the water column
$\begin{aligned} = 2 (A h ρ) g \frac{h}{2} - [A h_{1} ρ g \frac{h_{1}}{2} + A h_{2} ρ \frac{g h_{2}}{2}] \\ = A ρ g h^{2} - [A ρ g \frac{h_{1}^{2}}{2} + \frac{A ρ g h_{2}^{2}}{2}] \\ = A ρ g [{(\frac{h_{1} + h_{2}}{2})}^{2} - \frac{h_{1}^{2} + h_{2}^{2}}{2}] \\ = A ρ g {[\frac{h_{1} - h_{2}}{2}]}^{2} \end{aligned}$

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