Three forces acting on an object are given by F_01 = (-2.31 i + 7.25j) N, F_2 =

Question

Three forces acting on an object are given by F_01 = (-2.31 i + 7.25j) N, F_2 = (5.45i - 1.85j) N, and F_3 = (-49i) N. The object experiences an acceleration of magnitude 3.55 m/s^2. (a) What is the direction of the acceleration? degree (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 19.0 s? m/s (d) What are the velocity components of the object after 19.0 s? (Let the velocity be denoted by v.) v = (i + j) m/s

Explanation / Answer

a)Resultant force = F1+F2+F3

= -2.3i+7.25j+5.45i-1.85j-49i = -45.85i+5.4j N

Acceleration direction will be same as force direction = 180-tan-1(5.4/45.85) = 173.28o (counterclockwise from +x axis)

b)Magnitude of force = sqrt(45.852+5.42) = 46.17 N

Acceleration = 3.55m/s2

Mass of object = 46.17/3.55=13 kg

c)Speed after 19 seconds = 19*3.55 = 67.45 m/s

d) velocity components = 67.45*(-45.85i+5.4j)/sqrt(45.852+5.42) =  67.45*(-0.993i+0.117) = -66.98i+7.89j m/s

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