A force F= 13Ni +13Nj acts on the hockey puck. Determine the work done if the force results in the puck's displacement by 4.2m in the +x direction and the 25m in the -y direction.

Question

anonymous · Answer

what i did first was doing Pythagorean therefrom of the force and displacement. so F = 18.38N and m=4.89m. the y direction is actually 2.5m
so i use the work formula $$w=F \Delta r \cos \theta$$
W=18.83*4.89*cos(theta). 
i thought theta was 180 or 90, but i got it incorrect.

ash2326 · Answer

@igbona You don't have to find magnitude first, just the dot product of force and displacement will suffice. We have
$$F= (13i+13j )N$$
Displacement is 4.2 in x and 25 in -y
so
$$D= (4.2i-25j)m$$
Now 
$$Work =F.D$$
$$Work= (13i+13j).(4.2i-25j)$$
$$Work=(54.6-325 )Joules$$
$$Work= -270.4 Joules$$

anonymous · Answer

thank you. so , is the work formula just Work= F*D or Work= F*D*Cos(theta)? My insructor gave us F*D*cos(theta). I just want to make sure which formula is correct.