A body of mass 2 ~kg is acted upon by two forces each of magnitude 1 ~N and inclined at 60^ ^ ^ with each other. The acceleration of the body in m s is [ 60^ ^ ^ =0.5]

Correct answer: C. 0.75 Let us apply the parallelogram law of vectors to find the resultant of the forces. F_ net = F_ 1 ^ 2 +F_ 2 ^ 2 +2 × F_ 1 ~F_ 2 θ Given: F_ 1 =1, ~F_ 2 =1 and θ=60^ ^ ^ So, F_ net = (1)^ 2 +(1)^ 2 +2 × 1 × 1 × 60^ 0 = 1+1+2 × 1 / 2 = 3 ~N Now F=ma or, a=(F)/(m)= 3 2 m / s^ 2 Thus, the net acceleration of the body is 3 2 ~m / ^ 2

JEE Main

Exam: JEE Main
Subject: Physics
Chapter: Mathematics in Physics

JEE Main · Physics · Mathematics in Physics

A body of mass 2 ~kg is acted upon by two forces each of magnitude...

A body of mass

2 \mathrm{~kg}

is acted upon by two forces each of magnitude

1 \mathrm{~N}

and inclined at

60^{{}^{{}^{\circ}}}

with each other. The acceleration of the body in

\frac{\mathrm{m}}{\mathrm{s}}

\left[\cos 60^{{}^{{}^{\circ}}}=0.5\right]

Options

A $\sqrt{0.35}$
B $\sqrt{0.65}$
C $\sqrt{0.75}$
D $\sqrt{0.20}$

Correct answer

C. $\sqrt{0.75}$

Step-by-step solution

Let us apply the parallelogram law of vectors to find the resultant of the forces.

\mathrm{F}_{\text {net }}=\sqrt{\mathrm{F}_{1}^{2}+\mathrm{F}_{2}^{2}+2 \times \mathrm{F}_{1} \mathrm{~F}_{2} \cos \theta}

Given:

\mathrm{F}_{1}=1, \mathrm{~F}_{2}=1

and

\theta=60^{{}^{{}^{\circ}}}

So,

\mathrm{F}_{\text {net }}=\sqrt{(1)^{2}+(1)^{2}+2 \times 1 \times 1 \times \cos 60^{0}}=\sqrt{1+1+2 \times 1 / 2}=\sqrt{3} \mathrm{~N}

Now

\mathrm{F}=\mathrm{ma}

or,

a=\frac{F}{m}=\frac{\sqrt{3}}{2} m / s^{2}

Thus, the net acceleration of the body is

\frac{\sqrt{3}}{2} \mathrm{~m} / \mathrms^{2}

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