SATHEE: Kinematics Question 304

Kinematics Question 304

Question: A force $\vec{F} = 3 \hat{i} + c \hat{j} + 2 \hat{k}$ acting on a particle causes a displacement $\vec{S} = - 4 \hat{i} + 2 \hat{j} - 3 \hat{k}$ in it’s own direction. If the work done it’s 6J, then the value of c will be [DPMT 1997]

Options:

A) ${(A^{2} + B^{2} + \frac{A B}{\sqrt{3}})}^{1 / 2}$

B) $A + B$

C) ${(A^{2} + B^{2} + \sqrt{3} A B)}^{1 / 2}$

D) ${(A^{2} + B^{2} + A B)}^{1 / 2}$

Show Answer

Answer:

Correct Answer: D

Solution:

$| \vec{A} \times \vec{B} | = \sqrt{3} (\vec{A} . \vec{B})$

$A B \sin θ = \sqrt{3} A B \cos θ$
$\Rightarrow$ $\tan θ = \sqrt{3}$ $θ = 60^{\circ}$

Now $| \vec{R} | = | \vec{A} + \vec{B} | = \sqrt{A^{2} + B^{2} + 2 A B \cos θ}$

$= \sqrt{A^{2} + B^{2} + 2 A B (\frac{1}{2})}$

$= {(A^{2} + B^{2} + A B)}^{1 / 2}$

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