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PHXI06:WORK ENERGY AND POWER

355694 Work done in time $t$ on a body of mass $m$ which is acceleration from rest to speed $v$ in time $t_{1}$ as a function of time $t$ is given by:

1

\frac{1}{2} m v^{2} {(\frac{t}{t_{1}})}^{2}

2

\frac{1}{2} {(\frac{m v}{t_{1}})}^{2} t^{2}

3

m \frac{v}{t_{1}} t^{2}

4

\frac{1}{2} m \frac{v}{t_{1}} t^{2}

Explanation:

$a = v / t_{1}$ , Displacement after time $t$ :
$S = \frac{1}{2} a t^{2} = \frac{1}{2} \frac{v}{t_{1}} t^{2}$
$W = F S = m a S = m \frac{v}{t_{1}} \frac{1}{2} \frac{v}{t_{1}} t^{2} = \frac{1}{2} m v^{2} {(\frac{t}{t_{1}})}^{2}$

PHXI06:WORK ENERGY AND POWER

355695 A truck accelerates from speed $v$ to 2 $v$ . Work done in during this is

1 Three times as the work done in accelerating it from rest to

v

2 Same as the work done in accelerating it from rest to

v

3 Four times as the work done in accelerating it from rest to

v

4 Lless than the work done in accelerating it from rest to

v

Explanation:

Let $m$ be mass of the truck.
According to work-energy theorem
$W = K_{f} - K_{i}$
Work done in accelerating the truck from rest to speed $v$ is
$\begin{aligned} W_{1} = \frac{1}{2} m v^{2} - 0 (∵ u = 0) \\ = \frac{1}{2} m v^{2} \end{aligned}$
Work done in accelerating the truck from speed $v$ to 2 $v$ is
$W_{2} = \frac{1}{2} m (2 v)^{2} - \frac{1}{2} m (v)^{2}$
$= \frac{3}{2} m v^{2} = 3 \frac{1}{2} m v^{2} = 3 W_{1} (Using (1))$

KCET - 2013

PHXI06:WORK ENERGY AND POWER

355696 A pendulum of length 2 $m$ left at $A$ . When it reaches $B$ , it looses $10 %$ of its total energy due to air resistance. The velocity at $B$ is
supporting img

1 6

m / s

2 1

m / s

3 2

m / s

4 8

m / s

Explanation:

$\frac{1}{2} m v^{2} = (0.9) (m g h)$
$∴ v = \sqrt{1.8 g h}$
$= \sqrt{1.8 \times 10 \times 2}$
$= 6 m / s$

PHXI06:WORK ENERGY AND POWER

355694 Work done in time $t$ on a body of mass $m$ which is acceleration from rest to speed $v$ in time $t_{1}$ as a function of time $t$ is given by:

1

\frac{1}{2} m v^{2} {(\frac{t}{t_{1}})}^{2}

2

\frac{1}{2} {(\frac{m v}{t_{1}})}^{2} t^{2}

3

m \frac{v}{t_{1}} t^{2}

4

\frac{1}{2} m \frac{v}{t_{1}} t^{2}

Explanation:

$a = v / t_{1}$ , Displacement after time $t$ :
$S = \frac{1}{2} a t^{2} = \frac{1}{2} \frac{v}{t_{1}} t^{2}$
$W = F S = m a S = m \frac{v}{t_{1}} \frac{1}{2} \frac{v}{t_{1}} t^{2} = \frac{1}{2} m v^{2} {(\frac{t}{t_{1}})}^{2}$

PHXI06:WORK ENERGY AND POWER

355695 A truck accelerates from speed $v$ to 2 $v$ . Work done in during this is

1 Three times as the work done in accelerating it from rest to

v

2 Same as the work done in accelerating it from rest to

v

3 Four times as the work done in accelerating it from rest to

v

4 Lless than the work done in accelerating it from rest to

v

Explanation:

Let $m$ be mass of the truck.
According to work-energy theorem
$W = K_{f} - K_{i}$
Work done in accelerating the truck from rest to speed $v$ is
$\begin{aligned} W_{1} = \frac{1}{2} m v^{2} - 0 (∵ u = 0) \\ = \frac{1}{2} m v^{2} \end{aligned}$
Work done in accelerating the truck from speed $v$ to 2 $v$ is
$W_{2} = \frac{1}{2} m (2 v)^{2} - \frac{1}{2} m (v)^{2}$
$= \frac{3}{2} m v^{2} = 3 \frac{1}{2} m v^{2} = 3 W_{1} (Using (1))$

KCET - 2013

PHXI06:WORK ENERGY AND POWER

355696 A pendulum of length 2 $m$ left at $A$ . When it reaches $B$ , it looses $10 %$ of its total energy due to air resistance. The velocity at $B$ is
supporting img

1 6

m / s

2 1

m / s

3 2

m / s

4 8

m / s

Explanation:

$\frac{1}{2} m v^{2} = (0.9) (m g h)$
$∴ v = \sqrt{1.8 g h}$
$= \sqrt{1.8 \times 10 \times 2}$
$= 6 m / s$

PHXI06:WORK ENERGY AND POWER

355694 Work done in time $t$ on a body of mass $m$ which is acceleration from rest to speed $v$ in time $t_{1}$ as a function of time $t$ is given by:

1

\frac{1}{2} m v^{2} {(\frac{t}{t_{1}})}^{2}

2

\frac{1}{2} {(\frac{m v}{t_{1}})}^{2} t^{2}

3

m \frac{v}{t_{1}} t^{2}

4

\frac{1}{2} m \frac{v}{t_{1}} t^{2}

Explanation:

$a = v / t_{1}$ , Displacement after time $t$ :
$S = \frac{1}{2} a t^{2} = \frac{1}{2} \frac{v}{t_{1}} t^{2}$
$W = F S = m a S = m \frac{v}{t_{1}} \frac{1}{2} \frac{v}{t_{1}} t^{2} = \frac{1}{2} m v^{2} {(\frac{t}{t_{1}})}^{2}$

PHXI06:WORK ENERGY AND POWER

355695 A truck accelerates from speed $v$ to 2 $v$ . Work done in during this is

1 Three times as the work done in accelerating it from rest to

v

2 Same as the work done in accelerating it from rest to

v

3 Four times as the work done in accelerating it from rest to

v

4 Lless than the work done in accelerating it from rest to

v

Explanation:

Let $m$ be mass of the truck.
According to work-energy theorem
$W = K_{f} - K_{i}$
Work done in accelerating the truck from rest to speed $v$ is
$\begin{aligned} W_{1} = \frac{1}{2} m v^{2} - 0 (∵ u = 0) \\ = \frac{1}{2} m v^{2} \end{aligned}$
Work done in accelerating the truck from speed $v$ to 2 $v$ is
$W_{2} = \frac{1}{2} m (2 v)^{2} - \frac{1}{2} m (v)^{2}$
$= \frac{3}{2} m v^{2} = 3 \frac{1}{2} m v^{2} = 3 W_{1} (Using (1))$

KCET - 2013

PHXI06:WORK ENERGY AND POWER

355696 A pendulum of length 2 $m$ left at $A$ . When it reaches $B$ , it looses $10 %$ of its total energy due to air resistance. The velocity at $B$ is
supporting img

1 6

m / s

2 1

m / s

3 2

m / s

4 8

m / s

Explanation:

$\frac{1}{2} m v^{2} = (0.9) (m g h)$
$∴ v = \sqrt{1.8 g h}$
$= \sqrt{1.8 \times 10 \times 2}$
$= 6 m / s$

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