The figure here shows an overhead view of three horizontal forces acting on a cargo canister that was initially stationary but that now moves across a frictionless floor. The force magnitudes are F1 = 2.80 N, F2 = 3.60 N, and F3 = 10.0 N, and the indicated angles are θ2 = 52.0 ˚ and θ3 = 32.0 ˚. What is the net work done on the canister by the three forces during the first 4.00 m of displacement?
Answers
Answer:
Find the components of the net force.
Horizontally:
Fx = F3*cosΘ3 - F1 - F2*sinΘ2
Fx = 17.0N*cos31.0 - 3.40N - 4.40N*sin50.0 = 7.80 N
Vertically:
Fy = F3*sinΘ3 - F2*cosΘ2 = 17.0N*sin31.0 - 4.40N*cos50.0 = 5.93 N
|F| = √(Fx² + Fy²) = 9.80 N
net work = |F| * d = 39.2 J
Given:
F₁ = 2.80 N
F₂ =3.60 N
F₃ =10.0 N
θ₂= 52.0°
θ₃ = 32.0°
To Find:
The net work done on the canister by the 3 forces during the first 4.00 m of displacement.
Explanation:
Find the components of the net force.
Horizontal
F(x) = (F₃×cosθ₃) - F1 - (F₂×sinθ₂)
F(x)= (17.0N×cos31.0) - 3.40N - (4.40N×sin50.0)
= 7.80 N
Vertical
F(y) = (F₃×sinθ₃) - (F₂×cosθ₂)
= (17.0N×sin31.0) - (4.40N×cos50.0)
= 5.93 N
|F| = √(Fx² + Fy²)
= 9.80 N
Net work done= |F| × d
= 39.2 J
Answer is 39.2 J