Question

If a force F⃗ acts on an object as that object moves through a displacement s⃗ , the work done by that force equals the scalar product of F⃗ and s⃗ : W=F⃗ ⋅s⃗ . A certain object moves through displacement s⃗ =(4.00m)i^+(5.00m)j^. As it moves it is acted on by force F⃗ , which has x-component Fx = -12.0 N (1 N = 1 newton is the SI unit of force). The work done by this force is 26.0 N⋅m = 26.0 J (1 J = 1 joule = 1 newton-meter is the SI unit of work). Find the y-component of F⃗ .

Answer 1

Answer:

14.8 N

Step-by-step explanation:

Answer 2

Answer:

Fy=14.8N

Step-by-step explanation:

the dot product between $\vec{F}$ and $\vec{S}$

is given by:

W=$\vec{F}\cdot\vec{S}= S_xF_x+SyFy$

this gives the equation :

26= (4.00)(-12.0)+(5.00)Fy

26= -48+(5.00)Fy

adding 48 to both sides of the equation:

74= (5.00)Fy

dividing by 5, we get

Fy=74/5.00=14.8