Question

Three forces acting on an object are given by # = (-2.21 + 7.155) N, F2 = (4.559 – 2.55) N, and F3 = (-49.51) n. The object experiences an acceleration of magnitude 3.95 m/s2 (a) What is the direction of the acceleration? 5.6 x Note that the direction of the acceleration is the same as the direction of the net force. (counterclockwise from the +x-axis) (b) What is the mass of the object? 12.0 kg (c) If the object is initially at rest, what is its speed after 17.0 s? 67.15 m/s (d) What are the velocity components of the object after 17.0 s? (Let the velocity be denoted by v.) v= i+ 1 m/s

Answer 1

a) Fret = Ã TETA = -2.24+71595 +4.551 - 2:57 - 49.50 = -47.5 i +4.65Î o= tant uits 47.5 = 174.40 b) 1 Fret 1= ruzos)?+ 1.65) = 47.727N 1 4222 - 12:05 ry 47.727 may - Free Oce 3.95 Rom ...) speed at = 395x17 67.15 m/s component with Ta vcosol + V sino's = 67.15 cos 1744 +67.156in 174.4 = -66.83€ + 6.55 ſ mis Lokalitet 67.15 cos 17 E-book