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PHXI06:WORK ENERGY AND POWER

355757 A block of mass $M$ is kept on a plane from which it starts accelerating upwards from rest with a constant acceleration $a$ . During the time interval the work done by contact force on mass $M$ is
supporting img

1 Zero

2

\frac{1}{2} M a^{2} T^{2}

3

- M g a T^{2}

4

\frac{1}{2} M (g + a) a T^{2}

Explanation:

$W_{N} = N S$
$N = M g + M a$
$S = \frac{1}{2} a T^{2} \Rightarrow W_{N} = \frac{1}{2} M (a + g) a T^{2}$

PHXI06:WORK ENERGY AND POWER

355758 A cord is used to lower vertically a block of mass $M$ by a distance $d$ with constant downward acceleration $\frac{g}{4}$ . Work done by the cord on the block is

1

- 3 M g \frac{d}{4}

2

M g d

3

Mg \frac{d}{4}

4

3 M g \frac{d}{4}

Explanation:

When the block moves vertically downward with acceleration $\frac{g}{4}$ then tension in the cord
$T = M (g - \frac{g}{4}) = \frac{3}{4} M g$
Work done by the cord $\vec{F} \cdot \vec{S} = F S \cos θ$
$\begin{aligned} = T d \cos 180^{\circ} \\ = (- \frac{3}{4} M g) \times d = - 3 M g \frac{d}{4} . \end{aligned}$

PHXI06:WORK ENERGY AND POWER

355759 A block of mass $1 k g$ is pushed up a surface inclined to horizontal at an angle of $60^{\circ}$ by a force of $10 N$ parallel to the inclined surface as shown in figure. When the block is pushed up by $10 m$ along inclined surface, the work done against frictional force is
[ $g = 10 m / s^{2}$ ]
supporting img

1

5 \sqrt{3} J

2

5 J

3

5 \times 10^{3} J

4

10 J

Explanation:

As $f_{k} = μ_{k} N$
$∴ f_{k} = μ_{k} m g \cos θ$
$f_{k} = (0.1) (1) (10) \cos 60^{\circ}$
$f_{k} = 0.5 N$
Work done against frictional force is,
$W = f_{k} s \cos 180^{\circ} = 0.5 \times 10 \times (- 1)$
$W = - 5 J .$
supporting img

So correct option is (2).

JEE - 2024

PHXI06:WORK ENERGY AND POWER

355760 A force $F = 50 N$ is applied at one end of string, the other end of which is tied to a block of mass 10 $k g$ . The block is free to move on a frictionless horizontal surface. Take initial instant as $θ = 30^{\circ}$ and final instant as $37^{\circ}$ . For the time between these two instants, answer the following question Net work done by the force $F$ on the block is
supporting img

1

\frac{50}{3} J

2

75 J

3

\frac{100}{3} J

4 None of these

Explanation:

$l_{1} = \frac{2}{\sin 30^{\circ}} = 4 m, l_{2} = \frac{2}{\sin 37^{\circ}} = \frac{10}{3} m$
Displacement of point $P : l_{1} - l_{2} = 4 - \frac{10}{3} = \frac{2}{3} m$
supporting img
Work done, $W = F (l_{1} - l_{2}) = 50 \times \frac{2}{3} = \frac{100}{3} J$

PHXI06:WORK ENERGY AND POWER

355757 A block of mass $M$ is kept on a plane from which it starts accelerating upwards from rest with a constant acceleration $a$ . During the time interval the work done by contact force on mass $M$ is
supporting img

1 Zero

2

\frac{1}{2} M a^{2} T^{2}

3

- M g a T^{2}

4

\frac{1}{2} M (g + a) a T^{2}

Explanation:

$W_{N} = N S$
$N = M g + M a$
$S = \frac{1}{2} a T^{2} \Rightarrow W_{N} = \frac{1}{2} M (a + g) a T^{2}$

PHXI06:WORK ENERGY AND POWER

355758 A cord is used to lower vertically a block of mass $M$ by a distance $d$ with constant downward acceleration $\frac{g}{4}$ . Work done by the cord on the block is

1

- 3 M g \frac{d}{4}

2

M g d

3

Mg \frac{d}{4}

4

3 M g \frac{d}{4}

Explanation:

When the block moves vertically downward with acceleration $\frac{g}{4}$ then tension in the cord
$T = M (g - \frac{g}{4}) = \frac{3}{4} M g$
Work done by the cord $\vec{F} \cdot \vec{S} = F S \cos θ$
$\begin{aligned} = T d \cos 180^{\circ} \\ = (- \frac{3}{4} M g) \times d = - 3 M g \frac{d}{4} . \end{aligned}$

PHXI06:WORK ENERGY AND POWER

355759 A block of mass $1 k g$ is pushed up a surface inclined to horizontal at an angle of $60^{\circ}$ by a force of $10 N$ parallel to the inclined surface as shown in figure. When the block is pushed up by $10 m$ along inclined surface, the work done against frictional force is
[ $g = 10 m / s^{2}$ ]
supporting img

1

5 \sqrt{3} J

2

5 J

3

5 \times 10^{3} J

4

10 J

Explanation:

As $f_{k} = μ_{k} N$
$∴ f_{k} = μ_{k} m g \cos θ$
$f_{k} = (0.1) (1) (10) \cos 60^{\circ}$
$f_{k} = 0.5 N$
Work done against frictional force is,
$W = f_{k} s \cos 180^{\circ} = 0.5 \times 10 \times (- 1)$
$W = - 5 J .$
supporting img

So correct option is (2).

JEE - 2024

PHXI06:WORK ENERGY AND POWER

355760 A force $F = 50 N$ is applied at one end of string, the other end of which is tied to a block of mass 10 $k g$ . The block is free to move on a frictionless horizontal surface. Take initial instant as $θ = 30^{\circ}$ and final instant as $37^{\circ}$ . For the time between these two instants, answer the following question Net work done by the force $F$ on the block is
supporting img

1

\frac{50}{3} J

2

75 J

3

\frac{100}{3} J

4 None of these

Explanation:

$l_{1} = \frac{2}{\sin 30^{\circ}} = 4 m, l_{2} = \frac{2}{\sin 37^{\circ}} = \frac{10}{3} m$
Displacement of point $P : l_{1} - l_{2} = 4 - \frac{10}{3} = \frac{2}{3} m$
supporting img
Work done, $W = F (l_{1} - l_{2}) = 50 \times \frac{2}{3} = \frac{100}{3} J$

PHXI06:WORK ENERGY AND POWER

355757 A block of mass $M$ is kept on a plane from which it starts accelerating upwards from rest with a constant acceleration $a$ . During the time interval the work done by contact force on mass $M$ is
supporting img

1 Zero

2

\frac{1}{2} M a^{2} T^{2}

3

- M g a T^{2}

4

\frac{1}{2} M (g + a) a T^{2}

Explanation:

$W_{N} = N S$
$N = M g + M a$
$S = \frac{1}{2} a T^{2} \Rightarrow W_{N} = \frac{1}{2} M (a + g) a T^{2}$

PHXI06:WORK ENERGY AND POWER

355758 A cord is used to lower vertically a block of mass $M$ by a distance $d$ with constant downward acceleration $\frac{g}{4}$ . Work done by the cord on the block is

1

- 3 M g \frac{d}{4}

2

M g d

3

Mg \frac{d}{4}

4

3 M g \frac{d}{4}

Explanation:

When the block moves vertically downward with acceleration $\frac{g}{4}$ then tension in the cord
$T = M (g - \frac{g}{4}) = \frac{3}{4} M g$
Work done by the cord $\vec{F} \cdot \vec{S} = F S \cos θ$
$\begin{aligned} = T d \cos 180^{\circ} \\ = (- \frac{3}{4} M g) \times d = - 3 M g \frac{d}{4} . \end{aligned}$

PHXI06:WORK ENERGY AND POWER

355759 A block of mass $1 k g$ is pushed up a surface inclined to horizontal at an angle of $60^{\circ}$ by a force of $10 N$ parallel to the inclined surface as shown in figure. When the block is pushed up by $10 m$ along inclined surface, the work done against frictional force is
[ $g = 10 m / s^{2}$ ]
supporting img

1

5 \sqrt{3} J

2

5 J

3

5 \times 10^{3} J

4

10 J

Explanation:

As $f_{k} = μ_{k} N$
$∴ f_{k} = μ_{k} m g \cos θ$
$f_{k} = (0.1) (1) (10) \cos 60^{\circ}$
$f_{k} = 0.5 N$
Work done against frictional force is,
$W = f_{k} s \cos 180^{\circ} = 0.5 \times 10 \times (- 1)$
$W = - 5 J .$
supporting img

So correct option is (2).

JEE - 2024

PHXI06:WORK ENERGY AND POWER

355760 A force $F = 50 N$ is applied at one end of string, the other end of which is tied to a block of mass 10 $k g$ . The block is free to move on a frictionless horizontal surface. Take initial instant as $θ = 30^{\circ}$ and final instant as $37^{\circ}$ . For the time between these two instants, answer the following question Net work done by the force $F$ on the block is
supporting img

1

\frac{50}{3} J

2

75 J

3

\frac{100}{3} J

4 None of these

Explanation:

$l_{1} = \frac{2}{\sin 30^{\circ}} = 4 m, l_{2} = \frac{2}{\sin 37^{\circ}} = \frac{10}{3} m$
Displacement of point $P : l_{1} - l_{2} = 4 - \frac{10}{3} = \frac{2}{3} m$
supporting img
Work done, $W = F (l_{1} - l_{2}) = 50 \times \frac{2}{3} = \frac{100}{3} J$

PHXI06:WORK ENERGY AND POWER

355757 A block of mass $M$ is kept on a plane from which it starts accelerating upwards from rest with a constant acceleration $a$ . During the time interval the work done by contact force on mass $M$ is
supporting img

1 Zero

2

\frac{1}{2} M a^{2} T^{2}

3

- M g a T^{2}

4

\frac{1}{2} M (g + a) a T^{2}

Explanation:

$W_{N} = N S$
$N = M g + M a$
$S = \frac{1}{2} a T^{2} \Rightarrow W_{N} = \frac{1}{2} M (a + g) a T^{2}$

PHXI06:WORK ENERGY AND POWER

355758 A cord is used to lower vertically a block of mass $M$ by a distance $d$ with constant downward acceleration $\frac{g}{4}$ . Work done by the cord on the block is

1

- 3 M g \frac{d}{4}

2

M g d

3

Mg \frac{d}{4}

4

3 M g \frac{d}{4}

Explanation:

When the block moves vertically downward with acceleration $\frac{g}{4}$ then tension in the cord
$T = M (g - \frac{g}{4}) = \frac{3}{4} M g$
Work done by the cord $\vec{F} \cdot \vec{S} = F S \cos θ$
$\begin{aligned} = T d \cos 180^{\circ} \\ = (- \frac{3}{4} M g) \times d = - 3 M g \frac{d}{4} . \end{aligned}$

PHXI06:WORK ENERGY AND POWER

355759 A block of mass $1 k g$ is pushed up a surface inclined to horizontal at an angle of $60^{\circ}$ by a force of $10 N$ parallel to the inclined surface as shown in figure. When the block is pushed up by $10 m$ along inclined surface, the work done against frictional force is
[ $g = 10 m / s^{2}$ ]
supporting img

1

5 \sqrt{3} J

2

5 J

3

5 \times 10^{3} J

4

10 J

Explanation:

As $f_{k} = μ_{k} N$
$∴ f_{k} = μ_{k} m g \cos θ$
$f_{k} = (0.1) (1) (10) \cos 60^{\circ}$
$f_{k} = 0.5 N$
Work done against frictional force is,
$W = f_{k} s \cos 180^{\circ} = 0.5 \times 10 \times (- 1)$
$W = - 5 J .$
supporting img

So correct option is (2).

JEE - 2024

PHXI06:WORK ENERGY AND POWER

355760 A force $F = 50 N$ is applied at one end of string, the other end of which is tied to a block of mass 10 $k g$ . The block is free to move on a frictionless horizontal surface. Take initial instant as $θ = 30^{\circ}$ and final instant as $37^{\circ}$ . For the time between these two instants, answer the following question Net work done by the force $F$ on the block is
supporting img

1

\frac{50}{3} J

2

75 J

3

\frac{100}{3} J

4 None of these

Explanation:

$l_{1} = \frac{2}{\sin 30^{\circ}} = 4 m, l_{2} = \frac{2}{\sin 37^{\circ}} = \frac{10}{3} m$
Displacement of point $P : l_{1} - l_{2} = 4 - \frac{10}{3} = \frac{2}{3} m$
supporting img
Work done, $W = F (l_{1} - l_{2}) = 50 \times \frac{2}{3} = \frac{100}{3} J$

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