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PHXI06:WORK ENERGY AND POWER

355559 The resultant of two forces at right angle is 5 $N$ . When the angle between them is $120^{\circ}$ , the resultant is $\sqrt{13}$ . Then, the force is

1

\sqrt{20} N, \sqrt{5} N

2

\sqrt{12} N, \sqrt{13} N

3

\sqrt{40} N, \sqrt{15} N

4

3 N, 4 N

Explanation:

Let, $A$ and $B$ be the two forces.
$\sqrt{A^{2} + B^{2}} = 5$
or $A^{2} + B^{2} = 25 (1)$
and $A^{2} + B^{2} + 2 A B \cos 120^{\circ} = 13$
or $25 + 2 A B \times (- 1 / 2) = 13$
or $2 A B = 24 (2)$
Solving eq's (1) and (2), we get
$A = 3 N$
and $B = 4 N$

PHXI06:WORK ENERGY AND POWER

355560 Given $\vec{R} = \vec{A} + \vec{B}$ and $R = A = B$ . the angle between $\vec{A}$ and $\vec{B}$ is

1

90^{\circ}

2

60^{\circ}

3

180^{\circ}

4

120^{\circ}

Explanation:

For the resultant,
$\begin{aligned} R^{2} = R^{2} + R^{2} + 2 R^{2} \cos θ \\ \Rightarrow R^{2} = 2 R^{2} + 2 R^{2} \cos θ \\ \frac{1}{2} = 1 + \cos θ \end{aligned}$
or $\cos θ = - \frac{1}{2}$ or $θ = 120^{\circ}$

PHXI06:WORK ENERGY AND POWER

355561 Given that $\vec{A} + \vec{B} = \vec{C}$ and $\vec{C}$ is $⊥$ to $\vec{A}$ . Further if $| \vec{A} | = | \vec{C} |$ , then what is the angle between $\vec{A}$ and $\vec{B}$

1

\frac{π}{4}

radian

2

\frac{π}{2}

3

\frac{3 π}{4}

radian

4

π

radian

Explanation:

Here, $\vec{A} + \vec{B} = \vec{C}$ and $\vec{B} = \vec{C} - \vec{A}$
Since $\vec{C} ⊥ \vec{A}$ therefore
$B^{2} = C^{2} + A^{2} - 2 C A \cos (90^{\circ})$
Also, $| \vec{C} | = | \vec{A} |$
$B^{2} = 2 A^{2} \Rightarrow B = \sqrt{2} A$
Now, $A^{2} + B^{2} + 2 A B \cos θ = C^{2} = A^{2}$
Therefore we find $\cos θ = - \frac{1}{\sqrt{2}}$
This gives $θ = \frac{3 π}{4}$

PHXI06:WORK ENERGY AND POWER

355562 If $A B \cos θ = A B$ , then the angle between $A$ and $B$ is

1

0^{\circ}

2

45^{\circ}

3

90^{\circ}

4

180^{\circ}

Explanation:

As, $A B \cos θ = A B$
$\cos θ = 1 \Rightarrow θ = 0^{\circ}$

PHXI06:WORK ENERGY AND POWER

355559 The resultant of two forces at right angle is 5 $N$ . When the angle between them is $120^{\circ}$ , the resultant is $\sqrt{13}$ . Then, the force is

1

\sqrt{20} N, \sqrt{5} N

2

\sqrt{12} N, \sqrt{13} N

3

\sqrt{40} N, \sqrt{15} N

4

3 N, 4 N

Explanation:

Let, $A$ and $B$ be the two forces.
$\sqrt{A^{2} + B^{2}} = 5$
or $A^{2} + B^{2} = 25 (1)$
and $A^{2} + B^{2} + 2 A B \cos 120^{\circ} = 13$
or $25 + 2 A B \times (- 1 / 2) = 13$
or $2 A B = 24 (2)$
Solving eq's (1) and (2), we get
$A = 3 N$
and $B = 4 N$

PHXI06:WORK ENERGY AND POWER

355560 Given $\vec{R} = \vec{A} + \vec{B}$ and $R = A = B$ . the angle between $\vec{A}$ and $\vec{B}$ is

1

90^{\circ}

2

60^{\circ}

3

180^{\circ}

4

120^{\circ}

Explanation:

For the resultant,
$\begin{aligned} R^{2} = R^{2} + R^{2} + 2 R^{2} \cos θ \\ \Rightarrow R^{2} = 2 R^{2} + 2 R^{2} \cos θ \\ \frac{1}{2} = 1 + \cos θ \end{aligned}$
or $\cos θ = - \frac{1}{2}$ or $θ = 120^{\circ}$

PHXI06:WORK ENERGY AND POWER

355561 Given that $\vec{A} + \vec{B} = \vec{C}$ and $\vec{C}$ is $⊥$ to $\vec{A}$ . Further if $| \vec{A} | = | \vec{C} |$ , then what is the angle between $\vec{A}$ and $\vec{B}$

1

\frac{π}{4}

radian

2

\frac{π}{2}

3

\frac{3 π}{4}

radian

4

π

radian

Explanation:

Here, $\vec{A} + \vec{B} = \vec{C}$ and $\vec{B} = \vec{C} - \vec{A}$
Since $\vec{C} ⊥ \vec{A}$ therefore
$B^{2} = C^{2} + A^{2} - 2 C A \cos (90^{\circ})$
Also, $| \vec{C} | = | \vec{A} |$
$B^{2} = 2 A^{2} \Rightarrow B = \sqrt{2} A$
Now, $A^{2} + B^{2} + 2 A B \cos θ = C^{2} = A^{2}$
Therefore we find $\cos θ = - \frac{1}{\sqrt{2}}$
This gives $θ = \frac{3 π}{4}$

PHXI06:WORK ENERGY AND POWER

355562 If $A B \cos θ = A B$ , then the angle between $A$ and $B$ is

1

0^{\circ}

2

45^{\circ}

3

90^{\circ}

4

180^{\circ}

Explanation:

As, $A B \cos θ = A B$
$\cos θ = 1 \Rightarrow θ = 0^{\circ}$

PHXI06:WORK ENERGY AND POWER

355559 The resultant of two forces at right angle is 5 $N$ . When the angle between them is $120^{\circ}$ , the resultant is $\sqrt{13}$ . Then, the force is

1

\sqrt{20} N, \sqrt{5} N

2

\sqrt{12} N, \sqrt{13} N

3

\sqrt{40} N, \sqrt{15} N

4

3 N, 4 N

Explanation:

Let, $A$ and $B$ be the two forces.
$\sqrt{A^{2} + B^{2}} = 5$
or $A^{2} + B^{2} = 25 (1)$
and $A^{2} + B^{2} + 2 A B \cos 120^{\circ} = 13$
or $25 + 2 A B \times (- 1 / 2) = 13$
or $2 A B = 24 (2)$
Solving eq's (1) and (2), we get
$A = 3 N$
and $B = 4 N$

PHXI06:WORK ENERGY AND POWER

355560 Given $\vec{R} = \vec{A} + \vec{B}$ and $R = A = B$ . the angle between $\vec{A}$ and $\vec{B}$ is

1

90^{\circ}

2

60^{\circ}

3

180^{\circ}

4

120^{\circ}

Explanation:

For the resultant,
$\begin{aligned} R^{2} = R^{2} + R^{2} + 2 R^{2} \cos θ \\ \Rightarrow R^{2} = 2 R^{2} + 2 R^{2} \cos θ \\ \frac{1}{2} = 1 + \cos θ \end{aligned}$
or $\cos θ = - \frac{1}{2}$ or $θ = 120^{\circ}$

PHXI06:WORK ENERGY AND POWER

355561 Given that $\vec{A} + \vec{B} = \vec{C}$ and $\vec{C}$ is $⊥$ to $\vec{A}$ . Further if $| \vec{A} | = | \vec{C} |$ , then what is the angle between $\vec{A}$ and $\vec{B}$

1

\frac{π}{4}

radian

2

\frac{π}{2}

3

\frac{3 π}{4}

radian

4

π

radian

Explanation:

Here, $\vec{A} + \vec{B} = \vec{C}$ and $\vec{B} = \vec{C} - \vec{A}$
Since $\vec{C} ⊥ \vec{A}$ therefore
$B^{2} = C^{2} + A^{2} - 2 C A \cos (90^{\circ})$
Also, $| \vec{C} | = | \vec{A} |$
$B^{2} = 2 A^{2} \Rightarrow B = \sqrt{2} A$
Now, $A^{2} + B^{2} + 2 A B \cos θ = C^{2} = A^{2}$
Therefore we find $\cos θ = - \frac{1}{\sqrt{2}}$
This gives $θ = \frac{3 π}{4}$

PHXI06:WORK ENERGY AND POWER

355562 If $A B \cos θ = A B$ , then the angle between $A$ and $B$ is

1

0^{\circ}

2

45^{\circ}

3

90^{\circ}

4

180^{\circ}

Explanation:

As, $A B \cos θ = A B$
$\cos θ = 1 \Rightarrow θ = 0^{\circ}$

PHXI06:WORK ENERGY AND POWER

355559 The resultant of two forces at right angle is 5 $N$ . When the angle between them is $120^{\circ}$ , the resultant is $\sqrt{13}$ . Then, the force is

1

\sqrt{20} N, \sqrt{5} N

2

\sqrt{12} N, \sqrt{13} N

3

\sqrt{40} N, \sqrt{15} N

4

3 N, 4 N

Explanation:

Let, $A$ and $B$ be the two forces.
$\sqrt{A^{2} + B^{2}} = 5$
or $A^{2} + B^{2} = 25 (1)$
and $A^{2} + B^{2} + 2 A B \cos 120^{\circ} = 13$
or $25 + 2 A B \times (- 1 / 2) = 13$
or $2 A B = 24 (2)$
Solving eq's (1) and (2), we get
$A = 3 N$
and $B = 4 N$

PHXI06:WORK ENERGY AND POWER

355560 Given $\vec{R} = \vec{A} + \vec{B}$ and $R = A = B$ . the angle between $\vec{A}$ and $\vec{B}$ is

1

90^{\circ}

2

60^{\circ}

3

180^{\circ}

4

120^{\circ}

Explanation:

For the resultant,
$\begin{aligned} R^{2} = R^{2} + R^{2} + 2 R^{2} \cos θ \\ \Rightarrow R^{2} = 2 R^{2} + 2 R^{2} \cos θ \\ \frac{1}{2} = 1 + \cos θ \end{aligned}$
or $\cos θ = - \frac{1}{2}$ or $θ = 120^{\circ}$

PHXI06:WORK ENERGY AND POWER

355561 Given that $\vec{A} + \vec{B} = \vec{C}$ and $\vec{C}$ is $⊥$ to $\vec{A}$ . Further if $| \vec{A} | = | \vec{C} |$ , then what is the angle between $\vec{A}$ and $\vec{B}$

1

\frac{π}{4}

radian

2

\frac{π}{2}

3

\frac{3 π}{4}

radian

4

π

radian

Explanation:

Here, $\vec{A} + \vec{B} = \vec{C}$ and $\vec{B} = \vec{C} - \vec{A}$
Since $\vec{C} ⊥ \vec{A}$ therefore
$B^{2} = C^{2} + A^{2} - 2 C A \cos (90^{\circ})$
Also, $| \vec{C} | = | \vec{A} |$
$B^{2} = 2 A^{2} \Rightarrow B = \sqrt{2} A$
Now, $A^{2} + B^{2} + 2 A B \cos θ = C^{2} = A^{2}$
Therefore we find $\cos θ = - \frac{1}{\sqrt{2}}$
This gives $θ = \frac{3 π}{4}$

PHXI06:WORK ENERGY AND POWER

355562 If $A B \cos θ = A B$ , then the angle between $A$ and $B$ is

1

0^{\circ}

2

45^{\circ}

3

90^{\circ}

4

180^{\circ}

Explanation:

As, $A B \cos θ = A B$
$\cos θ = 1 \Rightarrow θ = 0^{\circ}$

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