Question

five force f1 ,f2,f3,f4,f5 are acting on a particle of mass 2.0kg so that it is moving with 4 m/s2in east direction if f1 is removed then accerlaration becomes 7 m/s2in north then acceleration of the block if only f1

Answer 1

Given that when all forces are acting on the particle then its acceleration is 4 m/s^2 towards East

So now we can write

$\vec F_1 + \vec F_2 + \vec F_3 + \vec F_4 + \vec F_5 = ma[tex]</p><p>[tex]\vec F_1 + \vec F_2 + \vec F_3 + \vec F_4 + \vec F_5 = 4*2 = 8\hat i[tex]</p><p>Now when force F1 is removed then </p><p>[tex]\vec F_2 + \vec F_3 + \vec F_4 + \vec F_5 = ma[tex]</p><p>[tex]\vec F_2 + \vec F_3 + \vec F_4 + \vec F_5 = 7*2 = 14 \hat j[tex]</p><p>Now subtract above two equations</p><p>[tex]\vec F_1 = 8\hat i - 14\hat j[tex]</p><p>now acceleration is given by</p><p>[tex]a_1 = \frac{F_1}{m}$

$a_1 = \frac{8\hat i - 14\hat j}{2}$

$a_1 = 4\hat i - 7\hat j$

so above is the acceleration of the mass when only F1 force is acting on it

Answer 2

Answer:

acceleration=$\sqrt{65}$

Explanation:

vector(F1+F2+F3+F4+F5) =2*(4i)=8i

Vector(F2+F3+F4+F5) =2*(7j)=14j

vector(F1) + 14j=8i

magnitude(F1)=$\sqrt{8^{2} +14^{2}\\ }$ = $\sqrt{260}$

a=F1/m=$\sqrt{65}$