A body of mass 5 kg under the action of constant force F-Fx -Fy - has velocity at t-0 s as u-6 -2 - m-s and at t-10 s as v-6- m-s- The force F is

Given u=6î 2ĵ v=6ĵ t=10 s By using, nbsp; v=u+at ⇒a=v ut a=8 ĵ 6 î10 a=4 ĵ5 3î5 Now F=ma (m=5 kg) =(4 ĵ 3 nbsp;î) N ...

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Standard XIII

Physics

Net Force

A body of mas...

Question

A body of mass $5 kg$ under the action of constant force $\to F = F_{x}^i + F_{y}^j$ has velocity at $t = 0 s$ as $\to u = (6^i - 2^j) m/s$ and at $t = 10 s$ as $\to v = 6^j m/s$ . The force $F$ is

(- 3^i + 4^j) N

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(- \frac{3}{5}^i + \frac{4}{5}^j) N

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(3^i - 4^j) N

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(\frac{3}{5}^i - \frac{4}{5}^j) N

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Solution

The correct option is A $(- 3^i + 4^j) N$
Given $\to u = 6^i - 2^j$
$\to v = 6^j$
$t = 10 s$
By using, $\to v = \to u + \to a t$
$\Rightarrow \to a = \frac{\to v - \to u}{t}$
$\to a = \frac{8^j - 6^i}{10}$
$\to a = \frac{4^j}{5} - \frac{3^i}{5}$
Now $\to F = m \to a$ $(m = 5 k g)$
= $(4^j - 3^i) N$

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A body of mass 5 kg under the action of constant force →F=Fx^i+Fy^j has velocity at t=0 s as →u=(6^i−2^j) m/s and at t=10 s as →v=6^j m/s. The force F is

The correct option is A (−3^i+4^j) NGiven →u=6^i−2^j →v=6^j t=10 s By using, →v=→u+→at ⇒→a=→v−→ut →a=8^j−6^i10 →a=4^j5−3^i5 Now →F=m→a (m=5 kg) =(4^j−3^i) N

A body of mass $5 kg$ under the action of constant force $\to F = F_{x}^i + F_{y}^j$ has velocity at $t = 0 s$ as $\to u = (6^i - 2^j) m/s$ and at $t = 10 s$ as $\to v = 6^j m/s$ . The force $F$ is

The correct option is A $(- 3^i + 4^j) N$
Given $\to u = 6^i - 2^j$
$\to v = 6^j$
$t = 10 s$
By using, $\to v = \to u + \to a t$
$\Rightarrow \to a = \frac{\to v - \to u}{t}$
$\to a = \frac{8^j - 6^i}{10}$
$\to a = \frac{4^j}{5} - \frac{3^i}{5}$
Now $\to F = m \to a$ $(m = 5 k g)$
= $(4^j - 3^i) N$