Three forces acting on an object are given by F vector_1 = (-1.5 i cap + 7.30 j

Question

Three forces acting on an object are given by F vector_1 = (-1.5 i cap + 7.30 j cap) N, F vector_2 = (4.60 i cap -1.5 j cap) N, and F vector_3 = (-48 i cap) N. The object experiences an acceleration of magnitude 3.95 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 20.0 s? m/s (d) What are the velocity components of the object after 20.0 s? (Let the velocity be denoted by v vector.) v vector = (i cap + j cap) m/s

Explanation / Answer

net force = F1 + F2 + F3

net force = -1.5i + 7.3j + 4.6i -1.5j -48i

net force = -44.9i + 5.8j

direction of force = direction of acceleration

direction of acceleration = tan^-1(5.8 / -44.9)

direction of acceleration = -7.36 degree or 360 - 7.36 counterclockwise

direction of acceleration = 352.64 degree counterclockwise

net force = sqrt(44.9^2 + 5.8^2)

net force = 45.2730604 N

force = mass * acceleration

45.2730604 = mass * 3.95

mass = 11.46 kg

by first equation of motion

v = u + at

v = 0 + 3.95 * 20

v = 79 m/s

speed after 20 sec = 79 m/s

horizontal component of velocity = -44.9 * 20 / 11.46

horizontal component of velocity = -78.359i m/s

vertical component of velocity = 5.8 * 20 / 11.46

vertical component of velocity = 10.12j m/s

v = (-78.359i + 10.12j) m/s

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