Answer:
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 ,
[tex]\sf \longrightarrow \boxed{\red{\rm F_{net}= \sqrt{F_1^2+F_2^2+2F_1F_2cos\theta }}}[/tex]
where [tex]\theta[/tex] is the angle between the forces .
- Since the forces act in opposite directions , the angle between them will be 180° .
[tex]\sf \longrightarrow F_{net}= \sqrt{ (8.3N)^2+(6.6N)^2+2(8.3)(6.6)(cos180^o)}\\ [/tex]
[tex]\sf \longrightarrow F_{net}= \sqrt{ (8.3N)^2+(6.6N)^2+2(8.3)(6.6)(-1)}\\ [/tex]
[tex]\sf \longrightarrow F_{net}= \sqrt{ (8.3N)^2+(6.6N)^2-2(8.3)(6.6)}\\ [/tex]
- RHS is in form of (a-b)² = a² + b² -2ab inside the square root .
[tex]\sf \longrightarrow F_{net}=\sqrt{( 8.3N - 6.6N)^2} \\ [/tex]
[tex]\sf \longrightarrow F_{net}= 8.3N - 6.6N \\ [/tex]
[tex]\sf \longrightarrow \red{ F_{net}= 1.7N} \\ [/tex]
Now we know that ,
[tex]\sf \longrightarrow F_{net}= mass * acceleration \\ [/tex]
[tex]\sf \longrightarrow 1.7N = 31kg * a\\ [/tex]
[tex]\sf \longrightarrow a =\dfrac{1.7N}{31kg}\\ [/tex]
[tex]\sf \longrightarrow \underline{\boxed{\bf accl^n = 0.054 m/s^2}} \\ [/tex]
- And the box will move in the direction of application of 8.3N force .
