Question

Two forces (10i+14j)N
and (6i +10j) N act upon a body of mass 8 kg. Then the acceleration of the body is​

Answer 1

Answer:

128i + 192j

Explanation:

Resultant force = (10+6)i + (14+10)j

= 16i + 24j

So acceleration = f/m

a = 8(16i + 24j)

=128i + 192j

This is only possible if both of the forces are acting in the same direction.

Answer 2

Given:

Two forces (10i + 14j) N and (6i + 10j) N act upon a body of mass 8 kg

To find:

Acceleration of the body.

Calculation:

Let net force be f.

$\vec{f} = \{10 \hat{i} + 14 \hat{j} \} + \{6 \hat{i} + 10 \hat{j} \}$

$= > \vec{f} = \{16 \hat{i} + 24 \hat{j} \}$

Acceleration is given as force/mass:

$= > \vec{a} = \dfrac{ \{16 \hat{i} + 24 \hat{j} \} }{8}$

$= > \vec{a} = \{2 \hat{i} + 3 \hat{j} \}$

Net magnitude of acceleration:

$= > | \vec{a}| = \sqrt{ {(2)}^{2} + {(3)}^{2} }$

$= > | \vec{a}| = \sqrt{ 4 + 9 }$

$= > | \vec{a}| = \sqrt{ 13} \: m {s}^{ - 2}$

So , final answer is:

$\boxed{ \blue{ \bold{ | \vec{a}| = \sqrt{ 13} \: m {s}^{ - 2} }}}$