Three forces acting on an object are given by F 1 = (1.75 [i] + 7.20 [j] ) N, F

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

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