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PHXI06:WORK ENERGY AND POWER

355664 A mass of 1 $k g$ is acted upon by a single force $F = (4 \hat{i} + 4 \hat{j}) N$ . Under this force it is displaced from $(0.0)$ to $(1 m, 1 m)$ . If initially the speed of the particle was 2 $m / s$ , its final speed should be

1 6

m / s

2 4.5

m / s

3 8

m / s

4 4

m / s

Explanation:

Work done = change in kinetic energy
$∴ \int_{(0.0)}^{(1.1)} (4 \hat{i} + 4 \hat{j}) \cdot (d x \hat{i} + d y \hat{j}) = \frac{1}{2} \times 1 \times (v_{f}^{2} - v_{i}^{2})$
or $8 = \frac{1}{2} (v_{f}^{2} - 4)$
or $v_{f} = 4.5 m / s$

PHXI06:WORK ENERGY AND POWER

355665 A block of mass $M$ is attached to the lower end of a vertical spring. The spring is hung from a ceiling and has force constant value $K$ . The mass is released from rest with the spring initially unstretched. The maximum extension produced in the length of the spring will be

1

M g / k

2

2 M g / k

3

4 M g / k

4

M g / 2 k

Explanation:

Let $x$ be the extension in the spring
Applying conservation of energy
$M g x - \frac{1}{2} k x^{2} = 0$
$\Rightarrow x = \frac{2 M g}{k}$

PHXI06:WORK ENERGY AND POWER

355666 Under the action of a force a 2 $k g$ body moves such that its position $x$ as a function of time $t$ is given by $x = \frac{t^{4}}{4} + 3$ . Then work done by the force in first two seconds is

1 6

J

2 10

J

3 7

J

4 64

J

Explanation:

The position of the body is $x = \frac{t^{4}}{4} + 3$
$\begin{aligned} \frac{d x}{d t} & = \frac{4 t^{3}}{4} + 0 \\ v & = t^{3} \\ W & = Δ k = k_{f} - k_{i} \\ = \frac{1}{2} (2) [{(2^{3})}^{2} - 0^{3}] = 64 J \end{aligned}$

PHXI06:WORK ENERGY AND POWER

355667 An athlete $m = 60 k g$ in the Olympic games
covers a distance of $100 m$ in $10 s$ . His kinetic energy can be estimated to be in the range:

1

200 J - 500 J

2

2 \times 10^{5} J - 3 \times 10^{5} J

3

20000 J - 50000 J

4

2000 J - 5000 J

Explanation:

Approximate mass $= 60 k g$
Approximate velocity $= 10 m s^{- 1}$
Approximate kinetic energy
$= \frac{1}{2} \times 60 \times 100 = 3000 J$
So kinetic energy range $= 2000 J$ to $5000 J$ .

PHXI06:WORK ENERGY AND POWER

355664 A mass of 1 $k g$ is acted upon by a single force $F = (4 \hat{i} + 4 \hat{j}) N$ . Under this force it is displaced from $(0.0)$ to $(1 m, 1 m)$ . If initially the speed of the particle was 2 $m / s$ , its final speed should be

1 6

m / s

2 4.5

m / s

3 8

m / s

4 4

m / s

Explanation:

Work done = change in kinetic energy
$∴ \int_{(0.0)}^{(1.1)} (4 \hat{i} + 4 \hat{j}) \cdot (d x \hat{i} + d y \hat{j}) = \frac{1}{2} \times 1 \times (v_{f}^{2} - v_{i}^{2})$
or $8 = \frac{1}{2} (v_{f}^{2} - 4)$
or $v_{f} = 4.5 m / s$

PHXI06:WORK ENERGY AND POWER

355665 A block of mass $M$ is attached to the lower end of a vertical spring. The spring is hung from a ceiling and has force constant value $K$ . The mass is released from rest with the spring initially unstretched. The maximum extension produced in the length of the spring will be

1

M g / k

2

2 M g / k

3

4 M g / k

4

M g / 2 k

Explanation:

Let $x$ be the extension in the spring
Applying conservation of energy
$M g x - \frac{1}{2} k x^{2} = 0$
$\Rightarrow x = \frac{2 M g}{k}$

PHXI06:WORK ENERGY AND POWER

355666 Under the action of a force a 2 $k g$ body moves such that its position $x$ as a function of time $t$ is given by $x = \frac{t^{4}}{4} + 3$ . Then work done by the force in first two seconds is

1 6

J

2 10

J

3 7

J

4 64

J

Explanation:

The position of the body is $x = \frac{t^{4}}{4} + 3$
$\begin{aligned} \frac{d x}{d t} & = \frac{4 t^{3}}{4} + 0 \\ v & = t^{3} \\ W & = Δ k = k_{f} - k_{i} \\ = \frac{1}{2} (2) [{(2^{3})}^{2} - 0^{3}] = 64 J \end{aligned}$

PHXI06:WORK ENERGY AND POWER

355667 An athlete $m = 60 k g$ in the Olympic games
covers a distance of $100 m$ in $10 s$ . His kinetic energy can be estimated to be in the range:

1

200 J - 500 J

2

2 \times 10^{5} J - 3 \times 10^{5} J

3

20000 J - 50000 J

4

2000 J - 5000 J

Explanation:

Approximate mass $= 60 k g$
Approximate velocity $= 10 m s^{- 1}$
Approximate kinetic energy
$= \frac{1}{2} \times 60 \times 100 = 3000 J$
So kinetic energy range $= 2000 J$ to $5000 J$ .

PHXI06:WORK ENERGY AND POWER

355664 A mass of 1 $k g$ is acted upon by a single force $F = (4 \hat{i} + 4 \hat{j}) N$ . Under this force it is displaced from $(0.0)$ to $(1 m, 1 m)$ . If initially the speed of the particle was 2 $m / s$ , its final speed should be

1 6

m / s

2 4.5

m / s

3 8

m / s

4 4

m / s

Explanation:

Work done = change in kinetic energy
$∴ \int_{(0.0)}^{(1.1)} (4 \hat{i} + 4 \hat{j}) \cdot (d x \hat{i} + d y \hat{j}) = \frac{1}{2} \times 1 \times (v_{f}^{2} - v_{i}^{2})$
or $8 = \frac{1}{2} (v_{f}^{2} - 4)$
or $v_{f} = 4.5 m / s$

PHXI06:WORK ENERGY AND POWER

355665 A block of mass $M$ is attached to the lower end of a vertical spring. The spring is hung from a ceiling and has force constant value $K$ . The mass is released from rest with the spring initially unstretched. The maximum extension produced in the length of the spring will be

1

M g / k

2

2 M g / k

3

4 M g / k

4

M g / 2 k

Explanation:

Let $x$ be the extension in the spring
Applying conservation of energy
$M g x - \frac{1}{2} k x^{2} = 0$
$\Rightarrow x = \frac{2 M g}{k}$

PHXI06:WORK ENERGY AND POWER

355666 Under the action of a force a 2 $k g$ body moves such that its position $x$ as a function of time $t$ is given by $x = \frac{t^{4}}{4} + 3$ . Then work done by the force in first two seconds is

1 6

J

2 10

J

3 7

J

4 64

J

Explanation:

The position of the body is $x = \frac{t^{4}}{4} + 3$
$\begin{aligned} \frac{d x}{d t} & = \frac{4 t^{3}}{4} + 0 \\ v & = t^{3} \\ W & = Δ k = k_{f} - k_{i} \\ = \frac{1}{2} (2) [{(2^{3})}^{2} - 0^{3}] = 64 J \end{aligned}$

PHXI06:WORK ENERGY AND POWER

355667 An athlete $m = 60 k g$ in the Olympic games
covers a distance of $100 m$ in $10 s$ . His kinetic energy can be estimated to be in the range:

1

200 J - 500 J

2

2 \times 10^{5} J - 3 \times 10^{5} J

3

20000 J - 50000 J

4

2000 J - 5000 J

Explanation:

Approximate mass $= 60 k g$
Approximate velocity $= 10 m s^{- 1}$
Approximate kinetic energy
$= \frac{1}{2} \times 60 \times 100 = 3000 J$
So kinetic energy range $= 2000 J$ to $5000 J$ .

PHXI06:WORK ENERGY AND POWER

355664 A mass of 1 $k g$ is acted upon by a single force $F = (4 \hat{i} + 4 \hat{j}) N$ . Under this force it is displaced from $(0.0)$ to $(1 m, 1 m)$ . If initially the speed of the particle was 2 $m / s$ , its final speed should be

1 6

m / s

2 4.5

m / s

3 8

m / s

4 4

m / s

Explanation:

Work done = change in kinetic energy
$∴ \int_{(0.0)}^{(1.1)} (4 \hat{i} + 4 \hat{j}) \cdot (d x \hat{i} + d y \hat{j}) = \frac{1}{2} \times 1 \times (v_{f}^{2} - v_{i}^{2})$
or $8 = \frac{1}{2} (v_{f}^{2} - 4)$
or $v_{f} = 4.5 m / s$

PHXI06:WORK ENERGY AND POWER

355665 A block of mass $M$ is attached to the lower end of a vertical spring. The spring is hung from a ceiling and has force constant value $K$ . The mass is released from rest with the spring initially unstretched. The maximum extension produced in the length of the spring will be

1

M g / k

2

2 M g / k

3

4 M g / k

4

M g / 2 k

Explanation:

Let $x$ be the extension in the spring
Applying conservation of energy
$M g x - \frac{1}{2} k x^{2} = 0$
$\Rightarrow x = \frac{2 M g}{k}$

PHXI06:WORK ENERGY AND POWER

355666 Under the action of a force a 2 $k g$ body moves such that its position $x$ as a function of time $t$ is given by $x = \frac{t^{4}}{4} + 3$ . Then work done by the force in first two seconds is

1 6

J

2 10

J

3 7

J

4 64

J

Explanation:

The position of the body is $x = \frac{t^{4}}{4} + 3$
$\begin{aligned} \frac{d x}{d t} & = \frac{4 t^{3}}{4} + 0 \\ v & = t^{3} \\ W & = Δ k = k_{f} - k_{i} \\ = \frac{1}{2} (2) [{(2^{3})}^{2} - 0^{3}] = 64 J \end{aligned}$

PHXI06:WORK ENERGY AND POWER

355667 An athlete $m = 60 k g$ in the Olympic games
covers a distance of $100 m$ in $10 s$ . His kinetic energy can be estimated to be in the range:

1

200 J - 500 J

2

2 \times 10^{5} J - 3 \times 10^{5} J

3

20000 J - 50000 J

4

2000 J - 5000 J

Explanation:

Approximate mass $= 60 k g$
Approximate velocity $= 10 m s^{- 1}$
Approximate kinetic energy
$= \frac{1}{2} \times 60 \times 100 = 3000 J$
So kinetic energy range $= 2000 J$ to $5000 J$ .