A body of mass 5 kg is at rest- There force F1-10N due North F2-10N along East and F3-10-2N along N-W act on it simultaneously- The acceleration produced in the body is-

The correct option is B 4 ms^ 2 along NorthGiven, F⃗_⃗1⃗=10ĵ F⃗_⃗2⃗=10î F⃗_⃗3⃗=10√(2)[ i+j√(2)]= 10î+10ĵ Acceleration, a⃗=F⃗_⃗1⃗+F⃗_⃗2 ...

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Byju's Answer

Standard XII

Physics

Vector Addition

A body of mas...

Question

A body of mass $5 k g$ is at rest. There force $F_{1} = 10 N$ due North $F_{2} = 10 N$ along East and $F_{3} = 10 \sqrt{2} N$ along $N - W$ act on it simultaneously. The acceleration produced in the body is,

4 m s^{- 2}

along North

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2 m s^{- 2}

along North

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4 m s^{- 2}

along East

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2 m s^{- 2}

along East

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Solution

The correct option is B $4 m s^{- 2}$ along North
Given,

$_{1} = 10^j$

$_{2} = 10^i$

$_{3} = 10 \sqrt{2} [\frac{- i + j}{\sqrt{2}}] = - 10^i + 10^j$

Acceleration, $\to a = \frac{_{1} +_{2} +_{3}}{m} = \frac{10^j + 10^i - 10^i + 10^j}{5} = 4^j = 4 m s^{- 2}$ along North.

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A body of mass 5 kg is at rest. There force F1=10N due North F2=10N along East and F3=10√2N along N−W act on it simultaneously. The acceleration produced in the body is,

The correct option is B 4 ms−2 along NorthGiven,→F1=10^j→F2=10^i→F3=10√2[−i+j√2]=−10^i+10^jAcceleration, →a=→F1+→F2+→F3m=10^j+10^i−10^i+10^j5=4^j=4ms−2 along North.

A body of mass $5 k g$ is at rest. There force $F_{1} = 10 N$ due North $F_{2} = 10 N$ along East and $F_{3} = 10 \sqrt{2} N$ along $N - W$ act on it simultaneously. The acceleration produced in the body is,

The correct option is B $4 m s^{- 2}$ along North
Given,

$_{1} = 10^j$

$_{2} = 10^i$

$_{3} = 10 \sqrt{2} [\frac{- i + j}{\sqrt{2}}] = - 10^i + 10^j$

Acceleration, $\to a = \frac{_{1} +_{2} +_{3}}{m} = \frac{10^j + 10^i - 10^i + 10^j}{5} = 4^j = 4 m s^{- 2}$ along North.