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PHXI06:WORK ENERGY AND POWER

355808 A man starts walking from a point on the surface of earth (assumed smooth) and reaches diagonally opposite point. What is the work done by him?

1 Positive

2 Negative

3 Zero Said

4 Nothing can be

PHXI06:WORK ENERGY AND POWER

355809 The figure shows all the surfaces are frictionless and mass of the block, $m = 1 k g$ . The block and wedge are held initially at rest. Now wedge is given a horizontal acceleartion of $10 m / s^{2}$ by applying a force on the wedge, so that the block does not slip on the wedge. Then work done by the normal force in ground frame on the block in $\sqrt{3}$ seconds is
supporting img

1

100 \sqrt{3} J

2

30 J

3

150 J

4

60 J

Explanation:

Let $N$ be the normal force on the block
supporting img
$\begin{aligned} N \sin θ = m a \\ N \cos θ = m g \\ \Rightarrow & \tan θ = \frac{a}{g} = 1 \Rightarrow θ = 45^{\circ} \\ W & = (N \sin θ) (v t) = 150 J \end{aligned}$

PHXI06:WORK ENERGY AND POWER

355810 A force $\vec{F} = k [y \hat{i} + x \hat{j}]$ where $k$ is a positive constant acts on a particle moving in $x$ - $y$ plane starting from the point $(3, 5)$ , the particle is taken along a straight line to $(5, 7)$ . The work done by the force is:

1 15

K

2 Zero

3 35

K

4 20

K

Explanation:

$W = k [\int_{3, 5}^{5, 7} y d x + x d y] = \int_{3, 5}^{5, 7} d (x, y) = k [x y]_{3, 5}^{5, 7}$ $= k = [35 - 15] = 20 k$

PHXI06:WORK ENERGY AND POWER

355811 A particle moves under the effect of a force $F = c x^{2}$ from $x = 0$ to $x = x_{0}$ . The work done in the process is

1

\frac{c x_{0}^{3}}{3}

2

c x_{0}^{2}

3

c x_{0}^{3}

4

\frac{c x_{0}^{3}}{2}

Explanation:

$w = \int_{0}^{x_{0}} F d x = \int_{0}^{x_{0}} c x^{2} d x = \frac{c x_{0}^{3}}{3}$

PHXI06:WORK ENERGY AND POWER

355812 The displacement $x$ in metre of a particle of mass $m$ $k g$ moving in one dimension under the action of a force is related to the time $t$ in second by the equation $t = \sqrt{x} + 3$ , the work done by the force (in joule) in first six seconds is

1

9 m / 2

2 18

m

3 36

m

4 Zero

Explanation:

Here $t = \sqrt{x} + 3$
$\begin{aligned} x = (t - 3)^{2} = t^{2} - 6 t + 9 \\ v = \frac{d x}{d t} = 2 t - 6 \end{aligned}$
at $t = 0 s, v = 2 \times 0 - 6 = - 6$
at $t = 6 s, v = 2 \times 6 - 6 = + 6$
Initial and final $KE$ are same hence no work is done
$W = \frac{1}{2} m (v_{1}^{2} - v_{2}^{2}) = 0$

PHXI06:WORK ENERGY AND POWER

355808 A man starts walking from a point on the surface of earth (assumed smooth) and reaches diagonally opposite point. What is the work done by him?

1 Positive

2 Negative

3 Zero Said

4 Nothing can be

PHXI06:WORK ENERGY AND POWER

355809 The figure shows all the surfaces are frictionless and mass of the block, $m = 1 k g$ . The block and wedge are held initially at rest. Now wedge is given a horizontal acceleartion of $10 m / s^{2}$ by applying a force on the wedge, so that the block does not slip on the wedge. Then work done by the normal force in ground frame on the block in $\sqrt{3}$ seconds is
supporting img

1

100 \sqrt{3} J

2

30 J

3

150 J

4

60 J

Explanation:

Let $N$ be the normal force on the block
supporting img
$\begin{aligned} N \sin θ = m a \\ N \cos θ = m g \\ \Rightarrow & \tan θ = \frac{a}{g} = 1 \Rightarrow θ = 45^{\circ} \\ W & = (N \sin θ) (v t) = 150 J \end{aligned}$

PHXI06:WORK ENERGY AND POWER

355810 A force $\vec{F} = k [y \hat{i} + x \hat{j}]$ where $k$ is a positive constant acts on a particle moving in $x$ - $y$ plane starting from the point $(3, 5)$ , the particle is taken along a straight line to $(5, 7)$ . The work done by the force is:

1 15

K

2 Zero

3 35

K

4 20

K

Explanation:

$W = k [\int_{3, 5}^{5, 7} y d x + x d y] = \int_{3, 5}^{5, 7} d (x, y) = k [x y]_{3, 5}^{5, 7}$ $= k = [35 - 15] = 20 k$

PHXI06:WORK ENERGY AND POWER

355811 A particle moves under the effect of a force $F = c x^{2}$ from $x = 0$ to $x = x_{0}$ . The work done in the process is

1

\frac{c x_{0}^{3}}{3}

2

c x_{0}^{2}

3

c x_{0}^{3}

4

\frac{c x_{0}^{3}}{2}

Explanation:

$w = \int_{0}^{x_{0}} F d x = \int_{0}^{x_{0}} c x^{2} d x = \frac{c x_{0}^{3}}{3}$

PHXI06:WORK ENERGY AND POWER

355812 The displacement $x$ in metre of a particle of mass $m$ $k g$ moving in one dimension under the action of a force is related to the time $t$ in second by the equation $t = \sqrt{x} + 3$ , the work done by the force (in joule) in first six seconds is

1

9 m / 2

2 18

m

3 36

m

4 Zero

Explanation:

Here $t = \sqrt{x} + 3$
$\begin{aligned} x = (t - 3)^{2} = t^{2} - 6 t + 9 \\ v = \frac{d x}{d t} = 2 t - 6 \end{aligned}$
at $t = 0 s, v = 2 \times 0 - 6 = - 6$
at $t = 6 s, v = 2 \times 6 - 6 = + 6$
Initial and final $KE$ are same hence no work is done
$W = \frac{1}{2} m (v_{1}^{2} - v_{2}^{2}) = 0$

PHXI06:WORK ENERGY AND POWER

355808 A man starts walking from a point on the surface of earth (assumed smooth) and reaches diagonally opposite point. What is the work done by him?

1 Positive

2 Negative

3 Zero Said

4 Nothing can be

PHXI06:WORK ENERGY AND POWER

355809 The figure shows all the surfaces are frictionless and mass of the block, $m = 1 k g$ . The block and wedge are held initially at rest. Now wedge is given a horizontal acceleartion of $10 m / s^{2}$ by applying a force on the wedge, so that the block does not slip on the wedge. Then work done by the normal force in ground frame on the block in $\sqrt{3}$ seconds is
supporting img

1

100 \sqrt{3} J

2

30 J

3

150 J

4

60 J

Explanation:

Let $N$ be the normal force on the block
supporting img
$\begin{aligned} N \sin θ = m a \\ N \cos θ = m g \\ \Rightarrow & \tan θ = \frac{a}{g} = 1 \Rightarrow θ = 45^{\circ} \\ W & = (N \sin θ) (v t) = 150 J \end{aligned}$

PHXI06:WORK ENERGY AND POWER

355810 A force $\vec{F} = k [y \hat{i} + x \hat{j}]$ where $k$ is a positive constant acts on a particle moving in $x$ - $y$ plane starting from the point $(3, 5)$ , the particle is taken along a straight line to $(5, 7)$ . The work done by the force is:

1 15

K

2 Zero

3 35

K

4 20

K

Explanation:

$W = k [\int_{3, 5}^{5, 7} y d x + x d y] = \int_{3, 5}^{5, 7} d (x, y) = k [x y]_{3, 5}^{5, 7}$ $= k = [35 - 15] = 20 k$

PHXI06:WORK ENERGY AND POWER

355811 A particle moves under the effect of a force $F = c x^{2}$ from $x = 0$ to $x = x_{0}$ . The work done in the process is

1

\frac{c x_{0}^{3}}{3}

2

c x_{0}^{2}

3

c x_{0}^{3}

4

\frac{c x_{0}^{3}}{2}

Explanation:

$w = \int_{0}^{x_{0}} F d x = \int_{0}^{x_{0}} c x^{2} d x = \frac{c x_{0}^{3}}{3}$

PHXI06:WORK ENERGY AND POWER

355812 The displacement $x$ in metre of a particle of mass $m$ $k g$ moving in one dimension under the action of a force is related to the time $t$ in second by the equation $t = \sqrt{x} + 3$ , the work done by the force (in joule) in first six seconds is

1

9 m / 2

2 18

m

3 36

m

4 Zero

Explanation:

Here $t = \sqrt{x} + 3$
$\begin{aligned} x = (t - 3)^{2} = t^{2} - 6 t + 9 \\ v = \frac{d x}{d t} = 2 t - 6 \end{aligned}$
at $t = 0 s, v = 2 \times 0 - 6 = - 6$
at $t = 6 s, v = 2 \times 6 - 6 = + 6$
Initial and final $KE$ are same hence no work is done
$W = \frac{1}{2} m (v_{1}^{2} - v_{2}^{2}) = 0$

PHXI06:WORK ENERGY AND POWER

355808 A man starts walking from a point on the surface of earth (assumed smooth) and reaches diagonally opposite point. What is the work done by him?

1 Positive

2 Negative

3 Zero Said

4 Nothing can be

PHXI06:WORK ENERGY AND POWER

355809 The figure shows all the surfaces are frictionless and mass of the block, $m = 1 k g$ . The block and wedge are held initially at rest. Now wedge is given a horizontal acceleartion of $10 m / s^{2}$ by applying a force on the wedge, so that the block does not slip on the wedge. Then work done by the normal force in ground frame on the block in $\sqrt{3}$ seconds is
supporting img

1

100 \sqrt{3} J

2

30 J

3

150 J

4

60 J

Explanation:

Let $N$ be the normal force on the block
supporting img
$\begin{aligned} N \sin θ = m a \\ N \cos θ = m g \\ \Rightarrow & \tan θ = \frac{a}{g} = 1 \Rightarrow θ = 45^{\circ} \\ W & = (N \sin θ) (v t) = 150 J \end{aligned}$

PHXI06:WORK ENERGY AND POWER

355810 A force $\vec{F} = k [y \hat{i} + x \hat{j}]$ where $k$ is a positive constant acts on a particle moving in $x$ - $y$ plane starting from the point $(3, 5)$ , the particle is taken along a straight line to $(5, 7)$ . The work done by the force is:

1 15

K

2 Zero

3 35

K

4 20

K

Explanation:

$W = k [\int_{3, 5}^{5, 7} y d x + x d y] = \int_{3, 5}^{5, 7} d (x, y) = k [x y]_{3, 5}^{5, 7}$ $= k = [35 - 15] = 20 k$

PHXI06:WORK ENERGY AND POWER

355811 A particle moves under the effect of a force $F = c x^{2}$ from $x = 0$ to $x = x_{0}$ . The work done in the process is

1

\frac{c x_{0}^{3}}{3}

2

c x_{0}^{2}

3

c x_{0}^{3}

4

\frac{c x_{0}^{3}}{2}

Explanation:

$w = \int_{0}^{x_{0}} F d x = \int_{0}^{x_{0}} c x^{2} d x = \frac{c x_{0}^{3}}{3}$

PHXI06:WORK ENERGY AND POWER

355812 The displacement $x$ in metre of a particle of mass $m$ $k g$ moving in one dimension under the action of a force is related to the time $t$ in second by the equation $t = \sqrt{x} + 3$ , the work done by the force (in joule) in first six seconds is

1

9 m / 2

2 18

m

3 36

m

4 Zero

Explanation:

Here $t = \sqrt{x} + 3$
$\begin{aligned} x = (t - 3)^{2} = t^{2} - 6 t + 9 \\ v = \frac{d x}{d t} = 2 t - 6 \end{aligned}$
at $t = 0 s, v = 2 \times 0 - 6 = - 6$
at $t = 6 s, v = 2 \times 6 - 6 = + 6$
Initial and final $KE$ are same hence no work is done
$W = \frac{1}{2} m (v_{1}^{2} - v_{2}^{2}) = 0$

PHXI06:WORK ENERGY AND POWER

355808 A man starts walking from a point on the surface of earth (assumed smooth) and reaches diagonally opposite point. What is the work done by him?

1 Positive

2 Negative

3 Zero Said

4 Nothing can be

PHXI06:WORK ENERGY AND POWER

355809 The figure shows all the surfaces are frictionless and mass of the block, $m = 1 k g$ . The block and wedge are held initially at rest. Now wedge is given a horizontal acceleartion of $10 m / s^{2}$ by applying a force on the wedge, so that the block does not slip on the wedge. Then work done by the normal force in ground frame on the block in $\sqrt{3}$ seconds is
supporting img

1

100 \sqrt{3} J

2

30 J

3

150 J

4

60 J

Explanation:

Let $N$ be the normal force on the block
supporting img
$\begin{aligned} N \sin θ = m a \\ N \cos θ = m g \\ \Rightarrow & \tan θ = \frac{a}{g} = 1 \Rightarrow θ = 45^{\circ} \\ W & = (N \sin θ) (v t) = 150 J \end{aligned}$

PHXI06:WORK ENERGY AND POWER

355810 A force $\vec{F} = k [y \hat{i} + x \hat{j}]$ where $k$ is a positive constant acts on a particle moving in $x$ - $y$ plane starting from the point $(3, 5)$ , the particle is taken along a straight line to $(5, 7)$ . The work done by the force is:

1 15

K

2 Zero

3 35

K

4 20

K

Explanation:

$W = k [\int_{3, 5}^{5, 7} y d x + x d y] = \int_{3, 5}^{5, 7} d (x, y) = k [x y]_{3, 5}^{5, 7}$ $= k = [35 - 15] = 20 k$

PHXI06:WORK ENERGY AND POWER

355811 A particle moves under the effect of a force $F = c x^{2}$ from $x = 0$ to $x = x_{0}$ . The work done in the process is

1

\frac{c x_{0}^{3}}{3}

2

c x_{0}^{2}

3

c x_{0}^{3}

4

\frac{c x_{0}^{3}}{2}

Explanation:

$w = \int_{0}^{x_{0}} F d x = \int_{0}^{x_{0}} c x^{2} d x = \frac{c x_{0}^{3}}{3}$

PHXI06:WORK ENERGY AND POWER

355812 The displacement $x$ in metre of a particle of mass $m$ $k g$ moving in one dimension under the action of a force is related to the time $t$ in second by the equation $t = \sqrt{x} + 3$ , the work done by the force (in joule) in first six seconds is

1

9 m / 2

2 18

m

3 36

m

4 Zero

Explanation:

Here $t = \sqrt{x} + 3$
$\begin{aligned} x = (t - 3)^{2} = t^{2} - 6 t + 9 \\ v = \frac{d x}{d t} = 2 t - 6 \end{aligned}$
at $t = 0 s, v = 2 \times 0 - 6 = - 6$
at $t = 6 s, v = 2 \times 6 - 6 = + 6$
Initial and final $KE$ are same hence no work is done
$W = \frac{1}{2} m (v_{1}^{2} - v_{2}^{2}) = 0$

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