A 0-2 kg object at rest is subjected to a force 0-3 -0-4 - N- What is the velocity of the object in m-s after 6 s

Here, m=0.2 kg, u=0 (initial velocity); F = ( 0.3 î 0.4 ĵ ) (Given) a=Fm= ( 0.3 î 0.4 ĵ )0.2= ( 32î 2 ĵ ) We know, v=u+at nbsp;v=0+ ( 32î 2 ĵ ) × 6=9 î 12ĵ ...

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Standard XIII

Physics

Equations of Motion

A 0.2 kg obje...

Question

A $0.2 kg$ object at rest is subjected to a force $(0.3^i - 0.4^j) N$ . What is the velocity of the object (in $m / s$ ) after $6 s$

(9^i - 12^j)

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(8^i - 16^j)

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(12^i - 9^j)

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(16^i - 8^j)

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Solution

The correct option is A $(9^i - 12^j)$
Here, $m = 0.2 kg, u = 0$ (initial velocity);
$\to F = (0.3^i - 0.4^j)$ (Given)
$\to a = \frac{\to F}{m} = \frac{(0.3^i - 0.4^j)}{0.2} = (\frac{3}{2}^i - 2^j)$
We know, $\to v = \to u + \to a t$
$⟹ \to v = 0 + (\frac{3}{2}^i - 2^j) \times 6 = 9^i - 12^j$

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A 0.2 kg object at rest is subjected to a force (0.3^i−0.4^j) N. What is the velocity of the object (in m/s) after 6 s

The correct option is A (9^i−12^j)Here, m=0.2 kg,u=0 (initial velocity); →F=(0.3^i−0.4^j) (Given) →a=→Fm=(0.3^i−0.4^j)0.2=(32^i−2^j) We know, →v=→u+→at ⟹→v=0+(32^i−2^j)×6=9^i−12^j

A $0.2 kg$ object at rest is subjected to a force $(0.3^i - 0.4^j) N$ . What is the velocity of the object (in $m / s$ ) after $6 s$

The correct option is A $(9^i - 12^j)$
Here, $m = 0.2 kg, u = 0$ (initial velocity);
$\to F = (0.3^i - 0.4^j)$ (Given)
$\to a = \frac{\to F}{m} = \frac{(0.3^i - 0.4^j)}{0.2} = (\frac{3}{2}^i - 2^j)$
We know, $\to v = \to u + \to a t$
$⟹ \to v = 0 + (\frac{3}{2}^i - 2^j) \times 6 = 9^i - 12^j$