Physics: Two men pull on a 31-kg box with forces 8.3 N and 6.6 N in opposite directions. Find the resultant acceleration of the box and the direction in which the box moves. Please HURRY

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Two men pull on a 31-kg box with forces 8.3 N and

tantoine2704

04.02.2022

0.054m/s²

Explanation:

Here we are given that a 31kg block is being pulled by two men who apply forces of 8.3N and 6.6N in opposite directions .

So we know that ,

$\sf \longrightarrow \boxed{\red{\rm F_{net}= \sqrt{F_1^2+F_2^2+2F_1F_2cos\theta }}}$

where $\theta$ is the angle between the forces .

Since the forces act in opposite directions , the angle between them will be 180° .

$\sf \longrightarrow F_{net}= \sqrt{ (8.3N)^2+(6.6N)^2+2(8.3)(6.6)(cos180^o)}\\$

Value of cos 180° = -1

$\sf \longrightarrow F_{net}= \sqrt{ (8.3N)^2+(6.6N)^2+2(8.3)(6.6)(-1)}\\$

$\sf \longrightarrow F_{net}= \sqrt{ (8.3N)^2+(6.6N)^2-2(8.3)(6.6)}\\$

RHS is in form of (a-b)² = a² + b² -2ab inside the square root .

$\sf \longrightarrow F_{net}=\sqrt{( 8.3N - 6.6N)^2} \\$

$\sf \longrightarrow F_{net}= 8.3N - 6.6N \\$

$\sf \longrightarrow \red{ F_{net}= 1.7N} \\$

Now we know that ,

$\sf \longrightarrow F_{net}= mass * acceleration \\$

$\sf \longrightarrow 1.7N = 31kg * a\\$

$\sf \longrightarrow a =\dfrac{1.7N}{31kg}\\$

$\sf \longrightarrow \underline{\boxed{\bf accl^n = 0.054 m/s^2}} \\$

And the box will move in the direction of application of 8.3N force .

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