physics
posted by Anonymous .
Two forces,
1 = (3.85 − 2.85) N
and
2 = (2.95 − 3.65) N,
act on a particle of mass 2.10 kg that is initially at rest at coordinates
(−2.30 m, −3.60 m).
(a) What are the components of the particle's velocity at t = 11.8 s?
= m/s
(b) In what direction is the particle moving at t = 11.8 s?
° counterclockwise from the +xaxis
(c) What displacement does the particle undergo during the first 11.8 s?
Δ = m
(d) What are the coordinates of the particle at t = 11.8 s?
x = m
y = m

F1 = (x,y) = (3.85N,2.85N)
F2 = (x,y) = (2.95N,3.65N).
a. Fr=X+Yi=(3.85+2.95) + (2.85i3.65i)=
6.8  6.5i = 9.41N[316.3o] = Resultant
force.
2.10kg particle
Location: (x,y) = (2.30m,3.60m)
a(x) = Fx/mass = 6.8/2.1 = 3.24 m/s^2
Vx=Vox + a*t = 0 + 3.24*11.8= 38.21 m/s.
a(y) = Fy/mass = 6.5/2.1 = 3.10 m/s^2
Vy=Voy + a*t = 0  3.1*11.8= 36.52 m/s.
b. Tan A = Vy/Vx = 36.52/38.21=0.95587
A = 43.71o
c. d(x) = (Vx^2Vox^2)/2a =
(38.21^20)/6.48 = 225.3 m.
Dx = 225.3  (2.30) = 227.6 m. = Hor.
comp. of displacement.
d(y) = (Vy^2Voy^2)/2a =
(36.52^20)/6.2 = 215.1 m.
Dy = 215.1  (3.60) = 211.5 m. = Ver.
comp. of displacement.
Tan Ar = Dy/Dx = 211.5/227.6 = 0.92933
Ar = 42.9o
Displacement=Dx/cosA=227.6/cos42.9=310.7
m.
X = 225.3 m
Y = 215.1 m.
d. (d(x),d(y)) = (225.3m,215.1m)