B)What is the magnitude of the mass\'s acceleration vector, a ? 2 sig figs C)Wha
ID: 2014236 • Letter: B
Question
B)What is the magnitude of the mass's acceleration vector, a ? 2 sig figs
C)What angle does Fr make with the positive x axis? 2 sig figs
D)What is the direction of a? In other words, what angle does this vector make with respect to the positive x axis? 2 sig figs
E)How far (in meters) will the mass move in 5.0 s? 2 sig figs
F)What is the magnitude of the velocity vector of the block at t=5.0 seconds? 2 sig figs
G)In what direction is the mass moving at time t=5.0 seconds ? That is, what angle does the velocity vector make with respect to the positive x axis? 2 sig figs
Explanation / Answer
mass of the block is m = 2 kg The magnitudes of the forces are F1 = 4 N F2 = 6 N F3 = 8 N The angle made by the force F1 with positive X axis is 1 = 25o The angle made by the force F2 with positive X axis is 2 = 325o The angle made by the force F3 with positive X axis is 3 = 180o The X components of the force are F1x = F1 cos1 = (4 N)cos25o = 3.625 N F2x = F2 cos2 = (6 N)cos325o = 4.915 N F3x = F3 cos3 = (8 N)cos180o = -8 N The net X component of the force is Fx = F1x + F2x + F3x Fx = 3.625 + 4.915 - 8 Fx = 0.54 N The Y components of the force are F1y = F1 sin1 = (4 N)sin25o = 1.69 N F2y = F2 sin2 = (6 N)sin325o = -3.44 N F3y = F3 sin3 = (8 N)sin180o = 0 The net Y component of the force is Fy = F1y + F2y + F3y Fy = 1.69 + (-3.44) + 0 Fy = -1.75 N a) The magnitude of the resultant force is F = [(0.54)2+(-1.75)2] F = 1.83 N b) The magnitude of the acceleration of the mass is a = F/m a = 1.83 N/2 kg a = 0.96 m/s2 c) The direction of the resultant force is = tan-1(-1.75/0.54) = -72.85o = 360-72.85 = 287.15o with the positive X direction d) The direction of the acceleration is also in the direction of the force i.e. 287.15o with the positive X direction e) The distance travelled in t = 5 s s = ut + 0.5at2 s = 0 + 0.5(0.96 m/s2)(5 s)2 s = 12 m f) The velocity of the mass after t = 5 s is v = u + at v = 0 + (0.96 m/s2)(5 s) v = 4.8 m/s g) The mass will mo9ve in the same direction as the resultant force acts i.e. the velocity vectro will make angle 287.15o with the positive X direction = 3.625 N F2x = F2 cos2 = (6 N)cos325o = 4.915 N F3x = F3 cos3 = (8 N)cos180o = -8 N The net X component of the force is Fx = F1x + F2x + F3x Fx = 3.625 + 4.915 - 8 Fx = 0.54 N The Y components of the force are F1y = F1 sin1 = (4 N)sin25o = 1.69 N F2y = F2 sin2 = (6 N)sin325o = -3.44 N F3y = F3 sin3 = (8 N)sin180o = 0 The net Y component of the force is Fy = F1y + F2y + F3y Fy = 1.69 + (-3.44) + 0 Fy = -1.75 N a) The magnitude of the resultant force is F = [(0.54)2+(-1.75)2] F = 1.83 N b) The magnitude of the acceleration of the mass is a = F/m a = 1.83 N/2 kg a = 0.96 m/s2 c) The direction of the resultant force is = tan-1(-1.75/0.54) = -72.85o = 360-72.85 = 287.15o with the positive X direction d) The direction of the acceleration is also in the direction of the force i.e. 287.15o with the positive X direction e) The distance travelled in t = 5 s s = ut + 0.5at2 s = 0 + 0.5(0.96 m/s2)(5 s)2 s = 12 m f) The velocity of the mass after t = 5 s is v = u + at v = 0 + (0.96 m/s2)(5 s) v = 4.8 m/s g) The mass will mo9ve in the same direction as the resultant force acts i.e. the velocity vectro will make angle 287.15o with the positive X direction F2x = F2 cos2 = (6 N)cos325o = 4.915 N F3x = F3 cos3 = (8 N)cos180o = -8 N The net X component of the force is Fx = F1x + F2x + F3x Fx = 3.625 + 4.915 - 8 Fx = 0.54 N The Y components of the force are F1y = F1 sin1 = (4 N)sin25o = 1.69 N F2y = F2 sin2 = (6 N)sin325o = -3.44 N F3y = F3 sin3 = (8 N)sin180o = 0 The net Y component of the force is Fy = F1y + F2y + F3y Fy = 1.69 + (-3.44) + 0 Fy = -1.75 N a) The magnitude of the resultant force is F = [(0.54)2+(-1.75)2] F = 1.83 N F3x = F3 cos3 = (8 N)cos180o = -8 N The net X component of the force is Fx = F1x + F2x + F3x Fx = 3.625 + 4.915 - 8 Fx = 0.54 N The Y components of the force are F1y = F1 sin1 = (4 N)sin25o = 1.69 N F2y = F2 sin2 = (6 N)sin325o = -3.44 N F3y = F3 sin3 = (8 N)sin180o = 0 The net Y component of the force is Fy = F1y + F2y + F3y Fy = 1.69 + (-3.44) + 0 Fy = -1.75 N a) The magnitude of the resultant force is F = [(0.54)2+(-1.75)2] F = 1.83 N = -8 N The net X component of the force is Fx = F1x + F2x + F3x Fx = 3.625 + 4.915 - 8 Fx = 0.54 N The Y components of the force are F1y = F1 sin1 = (4 N)sin25o = 1.69 N F2y = F2 sin2 = (6 N)sin325o = -3.44 N F3y = F3 sin3 = (8 N)sin180o = 0 The net Y component of the force is Fy = F1y + F2y + F3y Fy = 1.69 + (-3.44) + 0 Fy = -1.75 N a) The magnitude of the resultant force is F = [(0.54)2+(-1.75)2] F = 1.83 N The Y components of the force are F1y = F1 sin1 = (4 N)sin25o = 1.69 N F2y = F2 sin2 = (6 N)sin325o = -3.44 N F3y = F3 sin3 = (8 N)sin180o = 0 The net Y component of the force is Fy = F1y + F2y + F3y Fy = 1.69 + (-3.44) + 0 Fy = -1.75 N a) The magnitude of the resultant force is F = [(0.54)2+(-1.75)2] F = 1.83 N = 1.69 N F2y = F2 sin2 = (6 N)sin325o = -3.44 N F3y = F3 sin3 = (8 N)sin180o = 0 The net Y component of the force is Fy = F1y + F2y + F3y Fy = 1.69 + (-3.44) + 0 Fy = -1.75 N a) The magnitude of the resultant force is F = [(0.54)2+(-1.75)2] F = 1.83 N F2y = F2 sin2 = (6 N)sin325o = -3.44 N F3y = F3 sin3 = (8 N)sin180o = 0 The net Y component of the force is Fy = F1y + F2y + F3y Fy = 1.69 + (-3.44) + 0 Fy = -1.75 N a) The magnitude of the resultant force is F = [(0.54)2+(-1.75)2] F = 1.83 N = -3.44 N F3y = F3 sin3 = (8 N)sin180o = 0 The net Y component of the force is Fy = F1y + F2y + F3y Fy = 1.69 + (-3.44) + 0 Fy = -1.75 N a) The magnitude of the resultant force is F = [(0.54)2+(-1.75)2] F = 1.83 N F3y = F3 sin3 = (8 N)sin180o = 0 The net Y component of the force is Fy = F1y + F2y + F3y Fy = 1.69 + (-3.44) + 0 Fy = -1.75 N a) The magnitude of the resultant force is F = [(0.54)2+(-1.75)2] F = 1.83 N = 0 The net Y component of the force is Fy = F1y + F2y + F3y Fy = 1.69 + (-3.44) + 0 Fy = -1.75 N a) The magnitude of the resultant force is F = [(0.54)2+(-1.75)2] F = 1.83 N b) The magnitude of the acceleration of the mass is a = F/m a = 1.83 N/2 kg a = 0.96 m/s2 c) The direction of the resultant force is = tan-1(-1.75/0.54) = -72.85o = 360-72.85 = 287.15o with the positive X direction d) The direction of the acceleration is also in the direction of the force i.e. 287.15o with the positive X direction e) The distance travelled in t = 5 s s = ut + 0.5at2 s = 0 + 0.5(0.96 m/s2)(5 s)2 s = 12 m f) The velocity of the mass after t = 5 s is v = u + at v = 0 + (0.96 m/s2)(5 s) v = 4.8 m/s g) The mass will mo9ve in the same direction as the resultant force acts i.e. the velocity vectro will make angle 287.15o with the positive X direction c) The direction of the resultant force is = tan-1(-1.75/0.54) = -72.85o = 360-72.85 = 287.15o with the positive X direction d) The direction of the acceleration is also in the direction of the force i.e. 287.15o with the positive X direction e) The distance travelled in t = 5 s s = ut + 0.5at2 s = 0 + 0.5(0.96 m/s2)(5 s)2 s = 12 m f) The velocity of the mass after t = 5 s is v = u + at v = 0 + (0.96 m/s2)(5 s) v = 4.8 m/s g) The mass will mo9ve in the same direction as the resultant force acts i.e. the velocity vectro will make angle 287.15o with the positive X direction g) The mass will mo9ve in the same direction as the resultant force acts i.e. the velocity vectro will make angle 287.15o with the positive X directionRelated Questions
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