Work and Kinetic Energy I need help finding the angle using the data and the vector components : A → 0.20 kg to 200g (40 degrees) B → 0.275 kg to 275 g (155 degrees) Determine the net force of the two known masses (magnitude - m/a (gravitational acceleration) and direction) by using vector components. Write each force vector in unit-vector notation, add the forces together, and then calculate the net force. Show all work in an organized manner. The angle should be measured from the positive x-axis, as usual. by + ay= cy : cy = 2.4 Wa = 1.50 i + 1.26 j bx + ax= cx : cx = -0.94 Wb = -2.44 i + 1.14 j Wa + Wb = Wc Wc = -0.94 i + 2.40 j |a| = (1.50)2+(1.26)2 = 1.96 |b| = (-2.44)2+(1.14)2 = 2.69 |b| = (-0.94)2+(2.40)2 = 2.58 tan-1(jy/xi) tan-1(2.40/-0.94) = net force A = 0.20 kg X 9.8 m/s^2 net force A = 1.96 N net Force B = 0.275 kg X 9.8 m/s^2 net Force B = 2.695 N net Force B = 0.263 kg X 9.8 m/s^2 net Force C = 2.577 N = 2.58N Net Force C = 2.58 N
University Physics 1 - Work and Kinetic Energy
I need help finding the angle using the data and the vector components :
A → 0.20 kg to 200g (40 degrees)
B → 0.275 kg to 275 g (155 degrees)
Determine the net force of the two known masses (magnitude - m/a (gravitational acceleration) and direction) by using vector components. Write each force vector in unit-vector notation, add the forces together, and then calculate the net force. Show all work in an organized manner. The angle should be measured from the positive x-axis, as usual.
by + ay= cy : cy = 2.4 Wa = 1.50 i + 1.26 j
bx + ax= cx : cx = -0.94 Wb = -2.44 i + 1.14 j
Wa + Wb = Wc Wc = -0.94 i + 2.40 j
|a| = (1.50)2+(1.26)2 = 1.96 |b| = (-2.44)2+(1.14)2 = 2.69
|b| = (-0.94)2+(2.40)2 = 2.58
tan-1(jy/xi)
tan-1(2.40/-0.94) =
net force A = 0.20 kg X 9.8 m/s^2
net force A = 1.96 N
net Force B = 0.275 kg X 9.8 m/s^2
net Force B = 2.695 N
net Force B = 0.263 kg X 9.8 m/s^2
net Force C = 2.577 N = 2.58N
Net Force C = 2.58 N
Angle = ?
Trending now
This is a popular solution!
Step by step
Solved in 3 steps with 3 images
