Question

#1 Net Force and Acceleration (15 Points) A 3.05 kg mass is subject to two forces, given by Fi = 2.72 N. -1.15 N, 2.41 N) and F2 = (2.14 N, -1.96 N, 3.26 N). (a) What is the vector valued net or total force on the mass? (b) What is the vector valued acceleration of the mass? (c) What is the magnitude of the acceleration of the mass?

Answer 1

< consider the two forces acting on a mass m 3.o5kg - is given as F = 2.72 î - 1.15 ] +2.40 RN F = (2014 î - 1-96þ +3.26 Ê )N Resultant force F = Fita Fe= (2.72 -1-15ý +2.412) +(2.14 -1-9697320 71 FRS (2.72 +2:14) â - (1.15+|196) +(2.41 +3.20) R = 4.86 î _ 3.11 î + 5.67 Ř It is the total face on mass 1. Zk ĵ +5.67 k 3.05 ST (b) Vector valued acceleration is given by the a 4.86 î - 3:11 ↑ +5.67 P = 4.86 î -3:11 - 3.05 3.05 3.05 a = (1.59 î - 1.01 ] +1.85 Å m/s (c) magnitude of the acceleration is given by lät= |(1.59%01+(105) = 16.9707 miš 10 - 2.640 mis WYNDINNITU