Three forces acting on an object are given by F 1 = (1.75 [i] + 7.20 [j] ) N, F
ID: 1338069 • Letter: T
Question
Three forces acting on an object are given by
F 1 = (1.75 [i] + 7.20 [j] ) N,
F 2 = (4.80 [i] 2.5 [j] ) N,
and
F 3 = (49.5 [i] ) N.
The object experiences an acceleration of magnitude4.00 m/s2.
(a) What is the direction of the acceleration?
° (counterclockwise from the +x-axis)
(b) What is the mass of the object?
kg
(c) If the object is initially at rest, what is its speed after 17.0 s?
m/s
(d) What are the velocity components of the object after 17.0 s? (Let the velocity be denoted by
v .
)
v = [i] + [j] m/s
Explanation / Answer
F 1 = (1.75 [i] + 7.20 [j] ) N,
F 2 = (4.80 [i] 2.5 [j] ) N,
F 3 = (49.5 [i] ) N.
Total force F = F1 + F2 + F3 = (-46.45 i + 4.7 j ) N
magnitude of force |F| = SQRT( 46.45^2 + 4.7^2) = 46.69
acceleration magnitude = 4/ms^2
mass of the object m = F/a = 46.69/4 = 11.67 kg
acceleration vector a^ = F/11.67 = (-46.45 i + 4.7 j )/11.67 = (-3.9 i + 0.40 j) m/s/s
direction of acceleration = y/x = arctan(-4.7/46.45) = -5.70 with +x axis
speed v = at = 4 x 17 = 68 m/s
Velocity v^ = a^ x t = (-3.9 i + 0.40 j) x 17 m/s
= (-66.66 i + 6.8 j) m/s