Question

A 2.30-kg object is moving along the x-axis at 1.90 m/s. As it passes the origin, two forces F⃗ 1 and F⃗ 2 are applied, both in the y-direction (plus or minus). The forces are applied for 2.80 s, after which the object is at x = 5.32 m, y = 11.8 m. . i tried 0i-5.077j that is not correct nor is just -5j express anwser in Fx2,please.

If F⃗ 1=11.0j^NF→1=11.0j^N, what's F⃗ 2F→2?

Answer 1

m = 2.3 kg

initial velocity along x-axis =

initial velocity along y-axis =

t = 2.8 s

displacement along x-axis =

displacement along y-axis = .

let a = acceleration along y-axis.

Applying kinematic equation along y-axis :

=> a = 3.01 m/s/s.

Net force along y-axis : F = ma = 2.3*3.01 = 6.923 N.

Now, assuming both the forces in same direction i.e; +ve y-axis.

then, F1 + F2 = F

=> F2 = F - F1 = 6.923 - 11 = -4.077 N [-ve sign indicates that the direction of F2 is along -ve y-axis]

therefore, N. [answer]