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1st Equation of Motion

Three force ...

Question

Three force $_{1} = (2^i + 4^j) N :_{2} = (2^i - 4^j) N$ and $_{3} = (^k - 4^i - 2^j) N$ are applied on an object of mass $1 k g$ at rest at origin. The position of the object at $t = 2 s$ will be

(- 2 m, - 6 m)

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(- 4 m, 8 m)

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(3 m, 6 m)

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(2 m, - 3 m)

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Solution

The correct option is B $(- 4 m, 2 m)$

Step 1, The given data
Mass of the object (m) = 1 kg
Initially, it was at rest so $u = 0$
Step 2, Finding the acceleration
Net force is sum of all forces on the object
, $\to F = (2 + 2 - 4)^i + (4 - 4 - 2)^j + (0 + 0 + 1)^k = 0^i - 2^j +^k$
We know,
$\to F = m \to a$
So acceleration will be $\to a = \frac{\to F}{m} = \frac{0^i - 2^j +^k}{1 k g} N = (0^i - 2^j +^k) m / s^{2}$
Step 3, Finding the position
Position of the body in x-direction after 2 seconds
We know,
$S = u t + \frac{1}{2} a t^{2}$
The initial velocity is zero
So, $x = \frac{1}{2} a_{x} t^{2} = \frac{1}{2} a_{x} (4) = \frac{1}{2} \times 0 \times 4 = 0 m$
Position of the body in y-direction after 2 seconds
We know,
$S = u t + \frac{1}{2} a t^{2}$
Initial velocity is zero
So, $y = \frac{1}{2} a_{y} t^{2} = \frac{1}{2} a_{y} (4) = \frac{1}{2} \times - 2 \times 4 = - 4 m$
Position of the body in z-direction after 2 seconds
We know,
$S = u t + \frac{1}{2} a t^{2}$
Initial velocity is zero
So, $z = \frac{1}{2} a_{z} t^{2} = \frac{1}{2} a_{z} (4) = \frac{1}{2} \times 1 \times 4 = 2 m$
Hence the position of the body after 2 seconds will be $(0 m, - 4 m, 2 m)$
Hence the correct option is (B)

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Three force →F1=(2^i+4^j) N:→F2=(2^i−4^j) N and →F3=(^k−4^i−2^j) N are applied on an object of mass 1 kg at rest at origin. The position of the object at t=2s will be

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