Question

a boy of mass 5 kg is acted upon by two perpendicular force 8 Newton and 6 Newton give the magnitude and direction of the acceleration of the body​

Answer 1

Answer:

a = 2 m / s^2.

Explanation:

Since the applied forces are perpendicular to each other, angle between them is 90° .

Therefore,

$\implies \sf{ Resultant \: force} = \sqrt{8 {}^{2} + 6 {}^{2} + 2(8 \times 3 ) \cos(90 \degree) }N \\ \\ \implies \sf{ Resultant \: force} = \sqrt{8 {}^{2} + 6 {}^{2} + 2(8 \times 3 )0 }N \\ \\ \implies \sf{ Resultant \: force} = \sqrt{8 {}^{2} + 6 {}^{2} }N\\ \\ \implies \sf{ Resultant \: force} = \sqrt{64 + 36}N$

= > Resultant force = 10N

= > ma = 10N

= > 5 kg. a = 10N

= > a = 2 m/s^2

For direction :

= > tanA = 8 / 6 { A = angle between resultant force and body }

= > tanA = 4 / 3

= > A = $\tan^{-1}(\frac{4}{3})$ { shows the direction }

Answer 2

$\huge{\underline{\underline{\red{\mathfrak{AnSwEr :}}}}}$

$\small{\underline{\blue{\sf{Given :}}}}$

• Mass of Boy (m) = 5 kg
• Force1 = 8 N
• Force2 = 6 N

$\rule{200}{1}$

$\small{\underline{\green{\sf{Solution :}}}}$

As the force are perpendicular so the angle between them is 90° , And whereas the Formula for the Resultant By Vector addition is :

$\large \star{\boxed{\sf{Resultant \: = \: \sqrt{a^2 \: + \: b^2 \: + \: 2ab \cos \theta}}}}$

Where,

• a = 8 N
• b = 6 N

$\implies {\sf{Resultant \: = \: \sqrt{(8)^2 \: + \: (6)^2 \: + \: 2(8)(6) \cos(90)}}} \\ \\ \implies {\sf{Resultant \: = \: \sqrt{64 \: + \: 36 \: + \: 2(8)(6)(0)}}} \\ \\ \implies {\sf{Resultant \: = \: \sqrt{64 \: + \: 36}}} \\ \\ \implies {\sf{Resultant \: = \: \sqrt{100}}} \\ \\ \implies {\sf{Resultant \: = \: 100 \: N}}$

Now,

$\implies {\sf{ma \: = \: 10}} \\ \\ \implies {\sf{5(a) \: = \: 10}} \\ \\ \implies {\sf{a \: = \: \dfrac{10}{5}}} \\ \\ \implies {\sf{a \: = \: 2 \: ms^{-2}}}$

∴ Acceleration is 2 m/s²