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PHXI06:WORK ENERGY AND POWER

355747 Starting at rest, a $10 k g$ object is acted upon by only one force as indicated in figure. Then the total work done by the force is
supporting img

1

90 J

2

125 J

3

245 J

4

490 J

Explanation:

Using impulse-momentum theorem
$P_{f} = P_{i} + \int F d t$
$P = m v = 0 + 30 (2) - 10 \times 1$
$P = m v = 50$
$\Rightarrow K E$ gained $= \frac{P^{2}}{2 m}$
$= \frac{2500}{20}$
$= 125 J .$

PHXI06:WORK ENERGY AND POWER

355748 Assertion :
Surface between the blocks $A$ and $B$ is rough, work done by friction on block $B$ is always negative.
Reason :
Total work done by friction on both the blocks is always zero
supporting img

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

Frictional force always acts in backward direction on block B. So work done on B is negative. If there is a sliding between $A$ and $B$ then work done by friction on the system is non zero. So option (3) is correct.

PHXI06:WORK ENERGY AND POWER

355750 A block of mass 5 $k g$ slides down a rough inclined surface. The angle of inclination is $45^{\circ}$ . The coefficient of sliding friction is 0.20 . When the block slides 10 $c m$ , the work done on the block by force of friction is

1

- \frac{1}{\sqrt{2}} J

2 1

J

3

- \sqrt{2} J

4

- 1 J

Explanation:

$\begin{aligned} W_{f} = f \cos 180^{\circ} = (μ mg \cos θ) (- 1) (S) \\ = - 0.2 \times 5 \times 10 \times \frac{1}{\sqrt{2}} \times 0.1 \\ = - \frac{1}{\sqrt{2}} J \end{aligned}$

PHXI06:WORK ENERGY AND POWER

355751 A particle is displaced from a position $(2 \hat{i} - \hat{j} + \hat{k})$ another position $(3 \hat{i} + 2 \hat{j} - 2 \hat{k})$ under the action of force $(2 \hat{i} + \hat{j} - \hat{k})$ . The work done by the force in an arbitrary unit is

1 8

2 10

3 12

4 16

Explanation:

Work done, $W = F \cdot r$
$= (2 \hat{i} + \hat{j} - \hat{k}) \cdot [(3 \hat{i} + 2 \hat{j} - 2 \hat{k})$
$- (2 \hat{i} - \hat{j} + \hat{k})] (g i v e n)$
$= (2 \hat{i} + \hat{j} - \hat{k}) \cdot (\hat{i} + 3 \hat{j} - 3 \hat{k})$
$W = 2 + 3 + 3 = 8 u n i t$

PHXI06:WORK ENERGY AND POWER

355752 The work done in pulling up a block of wood weighing 2 $k N$ for a length of 10 $m$ on a smooth plane inclined at an angle of $15^{\circ}$ with the horizontal is $[\sin 15^{\circ} = 0.2588]$

1 8.91

k J

2 9.82

k J

3 4.39

k J

4 5.17

k J

Explanation:

$W = m g \sin θ \times s$
$= 2 \times 10^{3} \times \sin 15^{\circ} \times 10$
$= 5.17 k J$

PHXI06:WORK ENERGY AND POWER

355747 Starting at rest, a $10 k g$ object is acted upon by only one force as indicated in figure. Then the total work done by the force is
supporting img

1

90 J

2

125 J

3

245 J

4

490 J

Explanation:

Using impulse-momentum theorem
$P_{f} = P_{i} + \int F d t$
$P = m v = 0 + 30 (2) - 10 \times 1$
$P = m v = 50$
$\Rightarrow K E$ gained $= \frac{P^{2}}{2 m}$
$= \frac{2500}{20}$
$= 125 J .$

PHXI06:WORK ENERGY AND POWER

355748 Assertion :
Surface between the blocks $A$ and $B$ is rough, work done by friction on block $B$ is always negative.
Reason :
Total work done by friction on both the blocks is always zero
supporting img

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

Frictional force always acts in backward direction on block B. So work done on B is negative. If there is a sliding between $A$ and $B$ then work done by friction on the system is non zero. So option (3) is correct.

PHXI06:WORK ENERGY AND POWER

355750 A block of mass 5 $k g$ slides down a rough inclined surface. The angle of inclination is $45^{\circ}$ . The coefficient of sliding friction is 0.20 . When the block slides 10 $c m$ , the work done on the block by force of friction is

1

- \frac{1}{\sqrt{2}} J

2 1

J

3

- \sqrt{2} J

4

- 1 J

Explanation:

$\begin{aligned} W_{f} = f \cos 180^{\circ} = (μ mg \cos θ) (- 1) (S) \\ = - 0.2 \times 5 \times 10 \times \frac{1}{\sqrt{2}} \times 0.1 \\ = - \frac{1}{\sqrt{2}} J \end{aligned}$

PHXI06:WORK ENERGY AND POWER

355751 A particle is displaced from a position $(2 \hat{i} - \hat{j} + \hat{k})$ another position $(3 \hat{i} + 2 \hat{j} - 2 \hat{k})$ under the action of force $(2 \hat{i} + \hat{j} - \hat{k})$ . The work done by the force in an arbitrary unit is

1 8

2 10

3 12

4 16

Explanation:

Work done, $W = F \cdot r$
$= (2 \hat{i} + \hat{j} - \hat{k}) \cdot [(3 \hat{i} + 2 \hat{j} - 2 \hat{k})$
$- (2 \hat{i} - \hat{j} + \hat{k})] (g i v e n)$
$= (2 \hat{i} + \hat{j} - \hat{k}) \cdot (\hat{i} + 3 \hat{j} - 3 \hat{k})$
$W = 2 + 3 + 3 = 8 u n i t$

PHXI06:WORK ENERGY AND POWER

355752 The work done in pulling up a block of wood weighing 2 $k N$ for a length of 10 $m$ on a smooth plane inclined at an angle of $15^{\circ}$ with the horizontal is $[\sin 15^{\circ} = 0.2588]$

1 8.91

k J

2 9.82

k J

3 4.39

k J

4 5.17

k J

Explanation:

$W = m g \sin θ \times s$
$= 2 \times 10^{3} \times \sin 15^{\circ} \times 10$
$= 5.17 k J$

PHXI06:WORK ENERGY AND POWER

355747 Starting at rest, a $10 k g$ object is acted upon by only one force as indicated in figure. Then the total work done by the force is
supporting img

1

90 J

2

125 J

3

245 J

4

490 J

Explanation:

Using impulse-momentum theorem
$P_{f} = P_{i} + \int F d t$
$P = m v = 0 + 30 (2) - 10 \times 1$
$P = m v = 50$
$\Rightarrow K E$ gained $= \frac{P^{2}}{2 m}$
$= \frac{2500}{20}$
$= 125 J .$

PHXI06:WORK ENERGY AND POWER

355748 Assertion :
Surface between the blocks $A$ and $B$ is rough, work done by friction on block $B$ is always negative.
Reason :
Total work done by friction on both the blocks is always zero
supporting img

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

Frictional force always acts in backward direction on block B. So work done on B is negative. If there is a sliding between $A$ and $B$ then work done by friction on the system is non zero. So option (3) is correct.

PHXI06:WORK ENERGY AND POWER

355750 A block of mass 5 $k g$ slides down a rough inclined surface. The angle of inclination is $45^{\circ}$ . The coefficient of sliding friction is 0.20 . When the block slides 10 $c m$ , the work done on the block by force of friction is

1

- \frac{1}{\sqrt{2}} J

2 1

J

3

- \sqrt{2} J

4

- 1 J

Explanation:

$\begin{aligned} W_{f} = f \cos 180^{\circ} = (μ mg \cos θ) (- 1) (S) \\ = - 0.2 \times 5 \times 10 \times \frac{1}{\sqrt{2}} \times 0.1 \\ = - \frac{1}{\sqrt{2}} J \end{aligned}$

PHXI06:WORK ENERGY AND POWER

355751 A particle is displaced from a position $(2 \hat{i} - \hat{j} + \hat{k})$ another position $(3 \hat{i} + 2 \hat{j} - 2 \hat{k})$ under the action of force $(2 \hat{i} + \hat{j} - \hat{k})$ . The work done by the force in an arbitrary unit is

1 8

2 10

3 12

4 16

Explanation:

Work done, $W = F \cdot r$
$= (2 \hat{i} + \hat{j} - \hat{k}) \cdot [(3 \hat{i} + 2 \hat{j} - 2 \hat{k})$
$- (2 \hat{i} - \hat{j} + \hat{k})] (g i v e n)$
$= (2 \hat{i} + \hat{j} - \hat{k}) \cdot (\hat{i} + 3 \hat{j} - 3 \hat{k})$
$W = 2 + 3 + 3 = 8 u n i t$

PHXI06:WORK ENERGY AND POWER

355752 The work done in pulling up a block of wood weighing 2 $k N$ for a length of 10 $m$ on a smooth plane inclined at an angle of $15^{\circ}$ with the horizontal is $[\sin 15^{\circ} = 0.2588]$

1 8.91

k J

2 9.82

k J

3 4.39

k J

4 5.17

k J

Explanation:

$W = m g \sin θ \times s$
$= 2 \times 10^{3} \times \sin 15^{\circ} \times 10$
$= 5.17 k J$

PHXI06:WORK ENERGY AND POWER

355747 Starting at rest, a $10 k g$ object is acted upon by only one force as indicated in figure. Then the total work done by the force is
supporting img

1

90 J

2

125 J

3

245 J

4

490 J

Explanation:

Using impulse-momentum theorem
$P_{f} = P_{i} + \int F d t$
$P = m v = 0 + 30 (2) - 10 \times 1$
$P = m v = 50$
$\Rightarrow K E$ gained $= \frac{P^{2}}{2 m}$
$= \frac{2500}{20}$
$= 125 J .$

PHXI06:WORK ENERGY AND POWER

355748 Assertion :
Surface between the blocks $A$ and $B$ is rough, work done by friction on block $B$ is always negative.
Reason :
Total work done by friction on both the blocks is always zero
supporting img

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

Frictional force always acts in backward direction on block B. So work done on B is negative. If there is a sliding between $A$ and $B$ then work done by friction on the system is non zero. So option (3) is correct.

PHXI06:WORK ENERGY AND POWER

355750 A block of mass 5 $k g$ slides down a rough inclined surface. The angle of inclination is $45^{\circ}$ . The coefficient of sliding friction is 0.20 . When the block slides 10 $c m$ , the work done on the block by force of friction is

1

- \frac{1}{\sqrt{2}} J

2 1

J

3

- \sqrt{2} J

4

- 1 J

Explanation:

$\begin{aligned} W_{f} = f \cos 180^{\circ} = (μ mg \cos θ) (- 1) (S) \\ = - 0.2 \times 5 \times 10 \times \frac{1}{\sqrt{2}} \times 0.1 \\ = - \frac{1}{\sqrt{2}} J \end{aligned}$

PHXI06:WORK ENERGY AND POWER

355751 A particle is displaced from a position $(2 \hat{i} - \hat{j} + \hat{k})$ another position $(3 \hat{i} + 2 \hat{j} - 2 \hat{k})$ under the action of force $(2 \hat{i} + \hat{j} - \hat{k})$ . The work done by the force in an arbitrary unit is

1 8

2 10

3 12

4 16

Explanation:

Work done, $W = F \cdot r$
$= (2 \hat{i} + \hat{j} - \hat{k}) \cdot [(3 \hat{i} + 2 \hat{j} - 2 \hat{k})$
$- (2 \hat{i} - \hat{j} + \hat{k})] (g i v e n)$
$= (2 \hat{i} + \hat{j} - \hat{k}) \cdot (\hat{i} + 3 \hat{j} - 3 \hat{k})$
$W = 2 + 3 + 3 = 8 u n i t$

PHXI06:WORK ENERGY AND POWER

355752 The work done in pulling up a block of wood weighing 2 $k N$ for a length of 10 $m$ on a smooth plane inclined at an angle of $15^{\circ}$ with the horizontal is $[\sin 15^{\circ} = 0.2588]$

1 8.91

k J

2 9.82

k J

3 4.39

k J

4 5.17

k J

Explanation:

$W = m g \sin θ \times s$
$= 2 \times 10^{3} \times \sin 15^{\circ} \times 10$
$= 5.17 k J$

PHXI06:WORK ENERGY AND POWER

355747 Starting at rest, a $10 k g$ object is acted upon by only one force as indicated in figure. Then the total work done by the force is
supporting img

1

90 J

2

125 J

3

245 J

4

490 J

Explanation:

Using impulse-momentum theorem
$P_{f} = P_{i} + \int F d t$
$P = m v = 0 + 30 (2) - 10 \times 1$
$P = m v = 50$
$\Rightarrow K E$ gained $= \frac{P^{2}}{2 m}$
$= \frac{2500}{20}$
$= 125 J .$

PHXI06:WORK ENERGY AND POWER

355748 Assertion :
Surface between the blocks $A$ and $B$ is rough, work done by friction on block $B$ is always negative.
Reason :
Total work done by friction on both the blocks is always zero
supporting img

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

Frictional force always acts in backward direction on block B. So work done on B is negative. If there is a sliding between $A$ and $B$ then work done by friction on the system is non zero. So option (3) is correct.

PHXI06:WORK ENERGY AND POWER

355750 A block of mass 5 $k g$ slides down a rough inclined surface. The angle of inclination is $45^{\circ}$ . The coefficient of sliding friction is 0.20 . When the block slides 10 $c m$ , the work done on the block by force of friction is

1

- \frac{1}{\sqrt{2}} J

2 1

J

3

- \sqrt{2} J

4

- 1 J

Explanation:

$\begin{aligned} W_{f} = f \cos 180^{\circ} = (μ mg \cos θ) (- 1) (S) \\ = - 0.2 \times 5 \times 10 \times \frac{1}{\sqrt{2}} \times 0.1 \\ = - \frac{1}{\sqrt{2}} J \end{aligned}$

PHXI06:WORK ENERGY AND POWER

355751 A particle is displaced from a position $(2 \hat{i} - \hat{j} + \hat{k})$ another position $(3 \hat{i} + 2 \hat{j} - 2 \hat{k})$ under the action of force $(2 \hat{i} + \hat{j} - \hat{k})$ . The work done by the force in an arbitrary unit is

1 8

2 10

3 12

4 16

Explanation:

Work done, $W = F \cdot r$
$= (2 \hat{i} + \hat{j} - \hat{k}) \cdot [(3 \hat{i} + 2 \hat{j} - 2 \hat{k})$
$- (2 \hat{i} - \hat{j} + \hat{k})] (g i v e n)$
$= (2 \hat{i} + \hat{j} - \hat{k}) \cdot (\hat{i} + 3 \hat{j} - 3 \hat{k})$
$W = 2 + 3 + 3 = 8 u n i t$

PHXI06:WORK ENERGY AND POWER

355752 The work done in pulling up a block of wood weighing 2 $k N$ for a length of 10 $m$ on a smooth plane inclined at an angle of $15^{\circ}$ with the horizontal is $[\sin 15^{\circ} = 0.2588]$

1 8.91

k J

2 9.82

k J

3 4.39

k J

4 5.17

k J

Explanation:

$W = m g \sin θ \times s$
$= 2 \times 10^{3} \times \sin 15^{\circ} \times 10$
$= 5.17 k J$