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PHXI06:WORK ENERGY AND POWER

355834 A 10 $k g$ brick moves along an $x$ -axis. Its acceleration as a function of its position is shown in figure. What is the net work performed on the brick by the force causing the acceleration as the brick moves from $x = 0$ to $x = 8.0 m$ ?
supporting img

1 4

J

2 8

J

3 2

J

4 1

J

Explanation:

According to the graph the acceleration $a$ varies linearly with the coordinate $x$ . We may write $a = α x$ , where $α$ is the slope of the graph
$α = \frac{20}{8} = 2.5 m s^{- 1}$
The force on the brick is in the positive $x$ direction and according to Newton's second law, its magnitude is given by
$F = \frac{a}{m} = \frac{α}{m} x$
if $x_{f}$ is the final coordinate, the work done by the force is
$\begin{aligned} W & = \int_{0}^{x_{f}} F d x = \frac{α}{m} \int_{0}^{x_{f}} x d x \\ = \frac{α}{2 m} x_{f}^{2} = \frac{2.5}{2 \times 10} \times (8)^{2} = 8 J \end{aligned}$

PHXI06:WORK ENERGY AND POWER

355835 A particle of mass 6 $k g$ moves according to the law $x = 0.2 t^{2} + 0.02 t^{3}$ . Find the work done by the force in first 4 $s$ .

1 1.1231

J

2 2.6428

J

3 2.1324

J

4 1.62228

J

Explanation:

$W = \frac{1}{2} m (v_{f}^{2} - v_{i}^{2})$
$\begin{aligned} v & = \frac{d x}{d t} = 0.4 t - 0.06 t^{2} \\ v & = (0.4) 4 - 0.06 (4)^{2} \\ = 0.64 m / s \\ = \frac{1}{2} \times 6 [(0.64)^{2} - 0] = 1.6228 J \end{aligned}$

PHXI06:WORK ENERGY AND POWER

355836 When a rubber-band is stretched by a distance $x$ , it exerts a restoring force of magnitude $F = a x + b x^{2}$ where $a$ and $b$ are constants. The work done in stretching the unstrected rubber band by $L$ is

1

\frac{a L^{2}}{2} + \frac{b L^{3}}{3}

2

\frac{1}{2} (\frac{a L^{2}}{2} + \frac{b L^{3}}{3})

3

a L^{2} + b L^{3}

4

\frac{1}{2} (a L^{2} + b L^{3})

Explanation:

$F = a x + b x^{2}$
$d W = F d x$
$W = \int_{0}^{L} (a x + b x^{2}) d x$
$W = \frac{a L^{2}}{2} + \frac{b L^{3}}{3}$

PHXI06:WORK ENERGY AND POWER

355837 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 $m$ is
supporting img

1 225

J

2 200

J

3 400

J

4 175

J

Explanation:

supporting img
Work done $W =$ area under $F - S$ graph $=$ area of trapezium $A B C D +$ area of trapezium $C E F D$
$= \frac{1}{2} \times (10 + 15) \times 10 + \frac{1}{2} \times (10 + 20) \times 5$
$= 125 + 75 = 200 J$

KCET - 2009

PHXI06:WORK ENERGY AND POWER

355834 A 10 $k g$ brick moves along an $x$ -axis. Its acceleration as a function of its position is shown in figure. What is the net work performed on the brick by the force causing the acceleration as the brick moves from $x = 0$ to $x = 8.0 m$ ?
supporting img

1 4

J

2 8

J

3 2

J

4 1

J

Explanation:

According to the graph the acceleration $a$ varies linearly with the coordinate $x$ . We may write $a = α x$ , where $α$ is the slope of the graph
$α = \frac{20}{8} = 2.5 m s^{- 1}$
The force on the brick is in the positive $x$ direction and according to Newton's second law, its magnitude is given by
$F = \frac{a}{m} = \frac{α}{m} x$
if $x_{f}$ is the final coordinate, the work done by the force is
$\begin{aligned} W & = \int_{0}^{x_{f}} F d x = \frac{α}{m} \int_{0}^{x_{f}} x d x \\ = \frac{α}{2 m} x_{f}^{2} = \frac{2.5}{2 \times 10} \times (8)^{2} = 8 J \end{aligned}$

PHXI06:WORK ENERGY AND POWER

355835 A particle of mass 6 $k g$ moves according to the law $x = 0.2 t^{2} + 0.02 t^{3}$ . Find the work done by the force in first 4 $s$ .

1 1.1231

J

2 2.6428

J

3 2.1324

J

4 1.62228

J

Explanation:

$W = \frac{1}{2} m (v_{f}^{2} - v_{i}^{2})$
$\begin{aligned} v & = \frac{d x}{d t} = 0.4 t - 0.06 t^{2} \\ v & = (0.4) 4 - 0.06 (4)^{2} \\ = 0.64 m / s \\ = \frac{1}{2} \times 6 [(0.64)^{2} - 0] = 1.6228 J \end{aligned}$

PHXI06:WORK ENERGY AND POWER

355836 When a rubber-band is stretched by a distance $x$ , it exerts a restoring force of magnitude $F = a x + b x^{2}$ where $a$ and $b$ are constants. The work done in stretching the unstrected rubber band by $L$ is

1

\frac{a L^{2}}{2} + \frac{b L^{3}}{3}

2

\frac{1}{2} (\frac{a L^{2}}{2} + \frac{b L^{3}}{3})

3

a L^{2} + b L^{3}

4

\frac{1}{2} (a L^{2} + b L^{3})

Explanation:

$F = a x + b x^{2}$
$d W = F d x$
$W = \int_{0}^{L} (a x + b x^{2}) d x$
$W = \frac{a L^{2}}{2} + \frac{b L^{3}}{3}$

PHXI06:WORK ENERGY AND POWER

355837 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 $m$ is
supporting img

1 225

J

2 200

J

3 400

J

4 175

J

Explanation:

supporting img
Work done $W =$ area under $F - S$ graph $=$ area of trapezium $A B C D +$ area of trapezium $C E F D$
$= \frac{1}{2} \times (10 + 15) \times 10 + \frac{1}{2} \times (10 + 20) \times 5$
$= 125 + 75 = 200 J$

KCET - 2009

PHXI06:WORK ENERGY AND POWER

355834 A 10 $k g$ brick moves along an $x$ -axis. Its acceleration as a function of its position is shown in figure. What is the net work performed on the brick by the force causing the acceleration as the brick moves from $x = 0$ to $x = 8.0 m$ ?
supporting img

1 4

J

2 8

J

3 2

J

4 1

J

Explanation:

According to the graph the acceleration $a$ varies linearly with the coordinate $x$ . We may write $a = α x$ , where $α$ is the slope of the graph
$α = \frac{20}{8} = 2.5 m s^{- 1}$
The force on the brick is in the positive $x$ direction and according to Newton's second law, its magnitude is given by
$F = \frac{a}{m} = \frac{α}{m} x$
if $x_{f}$ is the final coordinate, the work done by the force is
$\begin{aligned} W & = \int_{0}^{x_{f}} F d x = \frac{α}{m} \int_{0}^{x_{f}} x d x \\ = \frac{α}{2 m} x_{f}^{2} = \frac{2.5}{2 \times 10} \times (8)^{2} = 8 J \end{aligned}$

PHXI06:WORK ENERGY AND POWER

355835 A particle of mass 6 $k g$ moves according to the law $x = 0.2 t^{2} + 0.02 t^{3}$ . Find the work done by the force in first 4 $s$ .

1 1.1231

J

2 2.6428

J

3 2.1324

J

4 1.62228

J

Explanation:

$W = \frac{1}{2} m (v_{f}^{2} - v_{i}^{2})$
$\begin{aligned} v & = \frac{d x}{d t} = 0.4 t - 0.06 t^{2} \\ v & = (0.4) 4 - 0.06 (4)^{2} \\ = 0.64 m / s \\ = \frac{1}{2} \times 6 [(0.64)^{2} - 0] = 1.6228 J \end{aligned}$

PHXI06:WORK ENERGY AND POWER

355836 When a rubber-band is stretched by a distance $x$ , it exerts a restoring force of magnitude $F = a x + b x^{2}$ where $a$ and $b$ are constants. The work done in stretching the unstrected rubber band by $L$ is

1

\frac{a L^{2}}{2} + \frac{b L^{3}}{3}

2

\frac{1}{2} (\frac{a L^{2}}{2} + \frac{b L^{3}}{3})

3

a L^{2} + b L^{3}

4

\frac{1}{2} (a L^{2} + b L^{3})

Explanation:

$F = a x + b x^{2}$
$d W = F d x$
$W = \int_{0}^{L} (a x + b x^{2}) d x$
$W = \frac{a L^{2}}{2} + \frac{b L^{3}}{3}$

PHXI06:WORK ENERGY AND POWER

355837 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 $m$ is
supporting img

1 225

J

2 200

J

3 400

J

4 175

J

Explanation:

supporting img
Work done $W =$ area under $F - S$ graph $=$ area of trapezium $A B C D +$ area of trapezium $C E F D$
$= \frac{1}{2} \times (10 + 15) \times 10 + \frac{1}{2} \times (10 + 20) \times 5$
$= 125 + 75 = 200 J$

KCET - 2009

PHXI06:WORK ENERGY AND POWER

355834 A 10 $k g$ brick moves along an $x$ -axis. Its acceleration as a function of its position is shown in figure. What is the net work performed on the brick by the force causing the acceleration as the brick moves from $x = 0$ to $x = 8.0 m$ ?
supporting img

1 4

J

2 8

J

3 2

J

4 1

J

Explanation:

According to the graph the acceleration $a$ varies linearly with the coordinate $x$ . We may write $a = α x$ , where $α$ is the slope of the graph
$α = \frac{20}{8} = 2.5 m s^{- 1}$
The force on the brick is in the positive $x$ direction and according to Newton's second law, its magnitude is given by
$F = \frac{a}{m} = \frac{α}{m} x$
if $x_{f}$ is the final coordinate, the work done by the force is
$\begin{aligned} W & = \int_{0}^{x_{f}} F d x = \frac{α}{m} \int_{0}^{x_{f}} x d x \\ = \frac{α}{2 m} x_{f}^{2} = \frac{2.5}{2 \times 10} \times (8)^{2} = 8 J \end{aligned}$

PHXI06:WORK ENERGY AND POWER

355835 A particle of mass 6 $k g$ moves according to the law $x = 0.2 t^{2} + 0.02 t^{3}$ . Find the work done by the force in first 4 $s$ .

1 1.1231

J

2 2.6428

J

3 2.1324

J

4 1.62228

J

Explanation:

$W = \frac{1}{2} m (v_{f}^{2} - v_{i}^{2})$
$\begin{aligned} v & = \frac{d x}{d t} = 0.4 t - 0.06 t^{2} \\ v & = (0.4) 4 - 0.06 (4)^{2} \\ = 0.64 m / s \\ = \frac{1}{2} \times 6 [(0.64)^{2} - 0] = 1.6228 J \end{aligned}$

PHXI06:WORK ENERGY AND POWER

355836 When a rubber-band is stretched by a distance $x$ , it exerts a restoring force of magnitude $F = a x + b x^{2}$ where $a$ and $b$ are constants. The work done in stretching the unstrected rubber band by $L$ is

1

\frac{a L^{2}}{2} + \frac{b L^{3}}{3}

2

\frac{1}{2} (\frac{a L^{2}}{2} + \frac{b L^{3}}{3})

3

a L^{2} + b L^{3}

4

\frac{1}{2} (a L^{2} + b L^{3})

Explanation:

$F = a x + b x^{2}$
$d W = F d x$
$W = \int_{0}^{L} (a x + b x^{2}) d x$
$W = \frac{a L^{2}}{2} + \frac{b L^{3}}{3}$

PHXI06:WORK ENERGY AND POWER

355837 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 $m$ is
supporting img

1 225

J

2 200

J

3 400

J

4 175

J

Explanation:

supporting img
Work done $W =$ area under $F - S$ graph $=$ area of trapezium $A B C D +$ area of trapezium $C E F D$
$= \frac{1}{2} \times (10 + 15) \times 10 + \frac{1}{2} \times (10 + 20) \times 5$
$= 125 + 75 = 200 J$

KCET - 2009

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