QBManager

Work, Energy and Power

148814 The displacement of a body of mass $4 kg$ varies with time as $s = (t^{2} + 2 t) m$ . Now the workdone by all the forces acting on the body during the time interval $t = 3 s$ and $t = 5 s$ is

1

80 J

2

320 J

3

240 J

4

160 J

Explanation:

D Given that,
Mass $(m) = 4 kg$
and $s = (t^{2} + 2 t) m$
Forces acting on the body during the time
Interval $t = 3 s$ to $t = 5 s$
Now,
$s = t^{2} + 2 t$
$v = \frac{dx}{dt} = 2 t + 2$
From Work- Energy theorem,
$W_{net} = Δ K . E$
$W_{net} = K_{f} - K_{i} = \frac{1}{2} m (v_{f}^{2} - v_{i}^{2})$
$W_{net} = \frac{1}{2} \times 4 [(2 \times 5 + 2)^{2} - (2 \times 3 + 2)^{2}]$
$W_{net} = \frac{1}{2} \times 4 \times 80$
$W_{net} = 160 J$

AP EAMCET-28.04.2017

Work, Energy and Power

148816 Consider a particle on which constant forces ${\vec{F}}_{1} = \hat{i} + 2 \hat{j} + 3 \hat{k}$ and ${\vec{F}}_{2} = 4 \hat{i} - 5 \hat{j} - 2 \hat{k} N$ act together resulting in a displacement from position ${\vec{r}}_{1} = 20 \hat{i} + 15 \hat{j} cm$ to ${\vec{r}}_{2} = 7 \hat{k} cm$ . The total work done on the particle is

1

- 0.48 J

2

+ 0.48 J

3

- 4.5 J

4

+ 4.8 J

Explanation:

A $F_{1} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) N$
$F_{2} = (4 \hat{i} - 5 \hat{j} - 2 \hat{k}) N$
$r_{1} = 20 \hat{i} + 15 \hat{j} cm$
$r_{2} = 7 \hat{k} cm$
Total force on particle, $F = F_{1} + F_{2}$
$= (\hat{i} + 2 \hat{j} + 3 \hat{k}) + (4 \hat{i} - 5 \hat{j} - 2 \hat{k})$
$= (5 \hat{i} - 3 \hat{j} + \hat{k}) N$
Displacement of particle, $s = {\vec{r}}_{2} - {\vec{r}}_{1}$
$= 7 \hat{k} - (20 \hat{i} + 15 \hat{j})$
$= (- 20 \hat{i} - 15 \hat{j} + 7 \hat{k}) cm$
We know that,
Work $(W) =$ F.s
$= (5 \hat{i} - 3 \hat{j} + \hat{k}) \cdot (- 20 \hat{i} - 15 \hat{j} + 7 \hat{k})$
$= (- 100 + 45 + 7) \times 10^{- 2}$
$= - 0.48 J$

TS EAMCET (Engg.)-2017

Work, Energy and Power

148817 A motor pump lifts 6 tonnes of water from a well of depth $25 m$ to the first floor of height 35 $m$ from the ground floor in 20 minutes. The power of the pump (in $kW$ ) is: $[g = 10 {ms}^{- 2}$ ]

1 3

2 12

3 1.5

4 6

Explanation:

A Given,
$g = 10 m / s^{2}$
$m = 6$ tonnes $= 6000 kg$
$h_{1} = 35 m$
$t = 20$ minutes $= 20 \times 60 s = 1200 s$ .
Depth $(h_{2}) = 25 m$
Total height $(h) = h_{1} + h_{2} = (35 + 25) = 60 m$
We know that,
$Power (P) = \frac{work}{time}$
$P = \frac{mgh}{t} [∵ work = mgh]$
$= \frac{6000 \times 10 \times 60}{1200} = 3000 W$
$= 3 kW$

Karnataka CET-2017

Work, Energy and Power

148818 Figure shows three forces applied to a trunk that moves leftward by $3 m$ over a smooth floor. The force magnitudes are $F_{1} = 5 N, F_{2} =$ $9 N$ and $F_{3} = 3 N$ . The net work done on the trunk by the three forces.

1

1.50 J

2

2.40 J

3

3.00 J

4

6.00 J

Explanation:

A Given, ${\vec{F}}_{1} = 5.0 N, {\vec{F}}_{2} = 9 N, {\vec{F}}_{3} = 3 N, \vec{d} = 3 m$
We know that, $W = \vec{F} \cdot \vec{d} = Fd \cos θ$
Work done in ${\vec{F}}_{1}$ direction-
$W_{1} = F_{1} d \cos θ_{1}$
$W_{1} = (5) (3) \cos 0^{\circ} = 15 J$
${\vec{F}}_{2}$ makes an angle $(θ_{2}) = (180^{\circ} - θ^{\circ}) = (180^{\circ} - 60^{\circ}) = 120^{\circ}$
Work done in ${\vec{F}}_{2}$ direction-
$W_{2} = F_{2} d \cos θ_{2}$
$W_{2} = (9) \times (3) \cos 120^{\circ} = - 13.5 J$
Work done in ${\vec{F}}_{3}$ direction-
$W_{3} = F_{3} d \cos θ_{3}$
$W_{3} = 3 \times 3 \times \cos 90^{\circ} = 0$
Net work done three forces-
$W = W_{1} + W_{2} + W_{3}$
$W = 15 - 13.5 + 0$
$W = 1.50 J$

AIIMS-2017

Work, Energy and Power

148814 The displacement of a body of mass $4 kg$ varies with time as $s = (t^{2} + 2 t) m$ . Now the workdone by all the forces acting on the body during the time interval $t = 3 s$ and $t = 5 s$ is

1

80 J

2

320 J

3

240 J

4

160 J

Explanation:

D Given that,
Mass $(m) = 4 kg$
and $s = (t^{2} + 2 t) m$
Forces acting on the body during the time
Interval $t = 3 s$ to $t = 5 s$
Now,
$s = t^{2} + 2 t$
$v = \frac{dx}{dt} = 2 t + 2$
From Work- Energy theorem,
$W_{net} = Δ K . E$
$W_{net} = K_{f} - K_{i} = \frac{1}{2} m (v_{f}^{2} - v_{i}^{2})$
$W_{net} = \frac{1}{2} \times 4 [(2 \times 5 + 2)^{2} - (2 \times 3 + 2)^{2}]$
$W_{net} = \frac{1}{2} \times 4 \times 80$
$W_{net} = 160 J$

AP EAMCET-28.04.2017

Work, Energy and Power

148816 Consider a particle on which constant forces ${\vec{F}}_{1} = \hat{i} + 2 \hat{j} + 3 \hat{k}$ and ${\vec{F}}_{2} = 4 \hat{i} - 5 \hat{j} - 2 \hat{k} N$ act together resulting in a displacement from position ${\vec{r}}_{1} = 20 \hat{i} + 15 \hat{j} cm$ to ${\vec{r}}_{2} = 7 \hat{k} cm$ . The total work done on the particle is

1

- 0.48 J

2

+ 0.48 J

3

- 4.5 J

4

+ 4.8 J

Explanation:

A $F_{1} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) N$
$F_{2} = (4 \hat{i} - 5 \hat{j} - 2 \hat{k}) N$
$r_{1} = 20 \hat{i} + 15 \hat{j} cm$
$r_{2} = 7 \hat{k} cm$
Total force on particle, $F = F_{1} + F_{2}$
$= (\hat{i} + 2 \hat{j} + 3 \hat{k}) + (4 \hat{i} - 5 \hat{j} - 2 \hat{k})$
$= (5 \hat{i} - 3 \hat{j} + \hat{k}) N$
Displacement of particle, $s = {\vec{r}}_{2} - {\vec{r}}_{1}$
$= 7 \hat{k} - (20 \hat{i} + 15 \hat{j})$
$= (- 20 \hat{i} - 15 \hat{j} + 7 \hat{k}) cm$
We know that,
Work $(W) =$ F.s
$= (5 \hat{i} - 3 \hat{j} + \hat{k}) \cdot (- 20 \hat{i} - 15 \hat{j} + 7 \hat{k})$
$= (- 100 + 45 + 7) \times 10^{- 2}$
$= - 0.48 J$

TS EAMCET (Engg.)-2017

Work, Energy and Power

148817 A motor pump lifts 6 tonnes of water from a well of depth $25 m$ to the first floor of height 35 $m$ from the ground floor in 20 minutes. The power of the pump (in $kW$ ) is: $[g = 10 {ms}^{- 2}$ ]

1 3

2 12

3 1.5

4 6

Explanation:

A Given,
$g = 10 m / s^{2}$
$m = 6$ tonnes $= 6000 kg$
$h_{1} = 35 m$
$t = 20$ minutes $= 20 \times 60 s = 1200 s$ .
Depth $(h_{2}) = 25 m$
Total height $(h) = h_{1} + h_{2} = (35 + 25) = 60 m$
We know that,
$Power (P) = \frac{work}{time}$
$P = \frac{mgh}{t} [∵ work = mgh]$
$= \frac{6000 \times 10 \times 60}{1200} = 3000 W$
$= 3 kW$

Karnataka CET-2017

Work, Energy and Power

148818 Figure shows three forces applied to a trunk that moves leftward by $3 m$ over a smooth floor. The force magnitudes are $F_{1} = 5 N, F_{2} =$ $9 N$ and $F_{3} = 3 N$ . The net work done on the trunk by the three forces.

1

1.50 J

2

2.40 J

3

3.00 J

4

6.00 J

Explanation:

A Given, ${\vec{F}}_{1} = 5.0 N, {\vec{F}}_{2} = 9 N, {\vec{F}}_{3} = 3 N, \vec{d} = 3 m$
We know that, $W = \vec{F} \cdot \vec{d} = Fd \cos θ$
Work done in ${\vec{F}}_{1}$ direction-
$W_{1} = F_{1} d \cos θ_{1}$
$W_{1} = (5) (3) \cos 0^{\circ} = 15 J$
${\vec{F}}_{2}$ makes an angle $(θ_{2}) = (180^{\circ} - θ^{\circ}) = (180^{\circ} - 60^{\circ}) = 120^{\circ}$
Work done in ${\vec{F}}_{2}$ direction-
$W_{2} = F_{2} d \cos θ_{2}$
$W_{2} = (9) \times (3) \cos 120^{\circ} = - 13.5 J$
Work done in ${\vec{F}}_{3}$ direction-
$W_{3} = F_{3} d \cos θ_{3}$
$W_{3} = 3 \times 3 \times \cos 90^{\circ} = 0$
Net work done three forces-
$W = W_{1} + W_{2} + W_{3}$
$W = 15 - 13.5 + 0$
$W = 1.50 J$

AIIMS-2017

Work, Energy and Power

148814 The displacement of a body of mass $4 kg$ varies with time as $s = (t^{2} + 2 t) m$ . Now the workdone by all the forces acting on the body during the time interval $t = 3 s$ and $t = 5 s$ is

1

80 J

2

320 J

3

240 J

4

160 J

Explanation:

D Given that,
Mass $(m) = 4 kg$
and $s = (t^{2} + 2 t) m$
Forces acting on the body during the time
Interval $t = 3 s$ to $t = 5 s$
Now,
$s = t^{2} + 2 t$
$v = \frac{dx}{dt} = 2 t + 2$
From Work- Energy theorem,
$W_{net} = Δ K . E$
$W_{net} = K_{f} - K_{i} = \frac{1}{2} m (v_{f}^{2} - v_{i}^{2})$
$W_{net} = \frac{1}{2} \times 4 [(2 \times 5 + 2)^{2} - (2 \times 3 + 2)^{2}]$
$W_{net} = \frac{1}{2} \times 4 \times 80$
$W_{net} = 160 J$

AP EAMCET-28.04.2017

Work, Energy and Power

148816 Consider a particle on which constant forces ${\vec{F}}_{1} = \hat{i} + 2 \hat{j} + 3 \hat{k}$ and ${\vec{F}}_{2} = 4 \hat{i} - 5 \hat{j} - 2 \hat{k} N$ act together resulting in a displacement from position ${\vec{r}}_{1} = 20 \hat{i} + 15 \hat{j} cm$ to ${\vec{r}}_{2} = 7 \hat{k} cm$ . The total work done on the particle is

1

- 0.48 J

2

+ 0.48 J

3

- 4.5 J

4

+ 4.8 J

Explanation:

A $F_{1} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) N$
$F_{2} = (4 \hat{i} - 5 \hat{j} - 2 \hat{k}) N$
$r_{1} = 20 \hat{i} + 15 \hat{j} cm$
$r_{2} = 7 \hat{k} cm$
Total force on particle, $F = F_{1} + F_{2}$
$= (\hat{i} + 2 \hat{j} + 3 \hat{k}) + (4 \hat{i} - 5 \hat{j} - 2 \hat{k})$
$= (5 \hat{i} - 3 \hat{j} + \hat{k}) N$
Displacement of particle, $s = {\vec{r}}_{2} - {\vec{r}}_{1}$
$= 7 \hat{k} - (20 \hat{i} + 15 \hat{j})$
$= (- 20 \hat{i} - 15 \hat{j} + 7 \hat{k}) cm$
We know that,
Work $(W) =$ F.s
$= (5 \hat{i} - 3 \hat{j} + \hat{k}) \cdot (- 20 \hat{i} - 15 \hat{j} + 7 \hat{k})$
$= (- 100 + 45 + 7) \times 10^{- 2}$
$= - 0.48 J$

TS EAMCET (Engg.)-2017

Work, Energy and Power

148817 A motor pump lifts 6 tonnes of water from a well of depth $25 m$ to the first floor of height 35 $m$ from the ground floor in 20 minutes. The power of the pump (in $kW$ ) is: $[g = 10 {ms}^{- 2}$ ]

1 3

2 12

3 1.5

4 6

Explanation:

A Given,
$g = 10 m / s^{2}$
$m = 6$ tonnes $= 6000 kg$
$h_{1} = 35 m$
$t = 20$ minutes $= 20 \times 60 s = 1200 s$ .
Depth $(h_{2}) = 25 m$
Total height $(h) = h_{1} + h_{2} = (35 + 25) = 60 m$
We know that,
$Power (P) = \frac{work}{time}$
$P = \frac{mgh}{t} [∵ work = mgh]$
$= \frac{6000 \times 10 \times 60}{1200} = 3000 W$
$= 3 kW$

Karnataka CET-2017

Work, Energy and Power

148818 Figure shows three forces applied to a trunk that moves leftward by $3 m$ over a smooth floor. The force magnitudes are $F_{1} = 5 N, F_{2} =$ $9 N$ and $F_{3} = 3 N$ . The net work done on the trunk by the three forces.

1

1.50 J

2

2.40 J

3

3.00 J

4

6.00 J

Explanation:

A Given, ${\vec{F}}_{1} = 5.0 N, {\vec{F}}_{2} = 9 N, {\vec{F}}_{3} = 3 N, \vec{d} = 3 m$
We know that, $W = \vec{F} \cdot \vec{d} = Fd \cos θ$
Work done in ${\vec{F}}_{1}$ direction-
$W_{1} = F_{1} d \cos θ_{1}$
$W_{1} = (5) (3) \cos 0^{\circ} = 15 J$
${\vec{F}}_{2}$ makes an angle $(θ_{2}) = (180^{\circ} - θ^{\circ}) = (180^{\circ} - 60^{\circ}) = 120^{\circ}$
Work done in ${\vec{F}}_{2}$ direction-
$W_{2} = F_{2} d \cos θ_{2}$
$W_{2} = (9) \times (3) \cos 120^{\circ} = - 13.5 J$
Work done in ${\vec{F}}_{3}$ direction-
$W_{3} = F_{3} d \cos θ_{3}$
$W_{3} = 3 \times 3 \times \cos 90^{\circ} = 0$
Net work done three forces-
$W = W_{1} + W_{2} + W_{3}$
$W = 15 - 13.5 + 0$
$W = 1.50 J$

AIIMS-2017

Work, Energy and Power

148814 The displacement of a body of mass $4 kg$ varies with time as $s = (t^{2} + 2 t) m$ . Now the workdone by all the forces acting on the body during the time interval $t = 3 s$ and $t = 5 s$ is

1

80 J

2

320 J

3

240 J

4

160 J

Explanation:

D Given that,
Mass $(m) = 4 kg$
and $s = (t^{2} + 2 t) m$
Forces acting on the body during the time
Interval $t = 3 s$ to $t = 5 s$
Now,
$s = t^{2} + 2 t$
$v = \frac{dx}{dt} = 2 t + 2$
From Work- Energy theorem,
$W_{net} = Δ K . E$
$W_{net} = K_{f} - K_{i} = \frac{1}{2} m (v_{f}^{2} - v_{i}^{2})$
$W_{net} = \frac{1}{2} \times 4 [(2 \times 5 + 2)^{2} - (2 \times 3 + 2)^{2}]$
$W_{net} = \frac{1}{2} \times 4 \times 80$
$W_{net} = 160 J$

AP EAMCET-28.04.2017

Work, Energy and Power

148816 Consider a particle on which constant forces ${\vec{F}}_{1} = \hat{i} + 2 \hat{j} + 3 \hat{k}$ and ${\vec{F}}_{2} = 4 \hat{i} - 5 \hat{j} - 2 \hat{k} N$ act together resulting in a displacement from position ${\vec{r}}_{1} = 20 \hat{i} + 15 \hat{j} cm$ to ${\vec{r}}_{2} = 7 \hat{k} cm$ . The total work done on the particle is

1

- 0.48 J

2

+ 0.48 J

3

- 4.5 J

4

+ 4.8 J

Explanation:

A $F_{1} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) N$
$F_{2} = (4 \hat{i} - 5 \hat{j} - 2 \hat{k}) N$
$r_{1} = 20 \hat{i} + 15 \hat{j} cm$
$r_{2} = 7 \hat{k} cm$
Total force on particle, $F = F_{1} + F_{2}$
$= (\hat{i} + 2 \hat{j} + 3 \hat{k}) + (4 \hat{i} - 5 \hat{j} - 2 \hat{k})$
$= (5 \hat{i} - 3 \hat{j} + \hat{k}) N$
Displacement of particle, $s = {\vec{r}}_{2} - {\vec{r}}_{1}$
$= 7 \hat{k} - (20 \hat{i} + 15 \hat{j})$
$= (- 20 \hat{i} - 15 \hat{j} + 7 \hat{k}) cm$
We know that,
Work $(W) =$ F.s
$= (5 \hat{i} - 3 \hat{j} + \hat{k}) \cdot (- 20 \hat{i} - 15 \hat{j} + 7 \hat{k})$
$= (- 100 + 45 + 7) \times 10^{- 2}$
$= - 0.48 J$

TS EAMCET (Engg.)-2017

Work, Energy and Power

148817 A motor pump lifts 6 tonnes of water from a well of depth $25 m$ to the first floor of height 35 $m$ from the ground floor in 20 minutes. The power of the pump (in $kW$ ) is: $[g = 10 {ms}^{- 2}$ ]

1 3

2 12

3 1.5

4 6

Explanation:

A Given,
$g = 10 m / s^{2}$
$m = 6$ tonnes $= 6000 kg$
$h_{1} = 35 m$
$t = 20$ minutes $= 20 \times 60 s = 1200 s$ .
Depth $(h_{2}) = 25 m$
Total height $(h) = h_{1} + h_{2} = (35 + 25) = 60 m$
We know that,
$Power (P) = \frac{work}{time}$
$P = \frac{mgh}{t} [∵ work = mgh]$
$= \frac{6000 \times 10 \times 60}{1200} = 3000 W$
$= 3 kW$

Karnataka CET-2017

Work, Energy and Power

148818 Figure shows three forces applied to a trunk that moves leftward by $3 m$ over a smooth floor. The force magnitudes are $F_{1} = 5 N, F_{2} =$ $9 N$ and $F_{3} = 3 N$ . The net work done on the trunk by the three forces.

1

1.50 J

2

2.40 J

3

3.00 J

4

6.00 J

Explanation:

A Given, ${\vec{F}}_{1} = 5.0 N, {\vec{F}}_{2} = 9 N, {\vec{F}}_{3} = 3 N, \vec{d} = 3 m$
We know that, $W = \vec{F} \cdot \vec{d} = Fd \cos θ$
Work done in ${\vec{F}}_{1}$ direction-
$W_{1} = F_{1} d \cos θ_{1}$
$W_{1} = (5) (3) \cos 0^{\circ} = 15 J$
${\vec{F}}_{2}$ makes an angle $(θ_{2}) = (180^{\circ} - θ^{\circ}) = (180^{\circ} - 60^{\circ}) = 120^{\circ}$
Work done in ${\vec{F}}_{2}$ direction-
$W_{2} = F_{2} d \cos θ_{2}$
$W_{2} = (9) \times (3) \cos 120^{\circ} = - 13.5 J$
Work done in ${\vec{F}}_{3}$ direction-
$W_{3} = F_{3} d \cos θ_{3}$
$W_{3} = 3 \times 3 \times \cos 90^{\circ} = 0$
Net work done three forces-
$W = W_{1} + W_{2} + W_{3}$
$W = 15 - 13.5 + 0$
$W = 1.50 J$

AIIMS-2017