Three forces acting on an object are given by F_1 = (-2.451 + 5.80j) N, F_2 = (5
ID: 1420872 • Letter: T
Question
Three forces acting on an object are given by F_1 = (-2.451 + 5.80j) N, F_2 = (5.251 - 1.25J) N, and F_3 = (-411) N. The object experiences an acceleration of magnitude 3.80 m/s^2. What is the direction of the acceleration? What is the mass of the object? You know the acceleration of the object and you can calculate the magnitude of the net force. How do you find the mass from this information? kg If the object is initially at rest, what is its speed after 15.0 s? What are the velocity components of the object after 15.0 s? (Let the velocity be denoted by v.)Explanation / Answer
three acting force
F1 = -2.45 i + 5.8 j N
F2 = 5.25 i - 1.25 j N
F3 = -41 i N
Net force on object F = F1 + F2 + F3
F = -2.45 i + 5.8 j + 5.25 i - 1.25 j - 41 i = - 38.2 i + 4.55 j
direction of acceleration is equal to direction of force
direction of acceleration = theta = arctan(- 4.55/38.2) = - 6.8 degrees = 180 - 6.8 = 173.2 degrees counterclockwise from +x axis.
magnitude of F = sqrt(38.2^2 + 4.55^2) = 38.47 N
mass of object m = F/a = 38.47/3.8 = 10.12 kg
c)
using equation
v = u + at
v = 0 + 3.8*15 = 57 m/s
d)
acceleration a = a*cos(theta) i + a*sin(theta) j = - 3.77 i + 0.45 j
vector v = a*t = ( - 56.55 i + 6.75 j ) m/s