Question

 

The diagram below shows a block of mass m = 2.00 kg on a frictionless horizontal surface, as seen from above. Three forces of magnitudes F1 = 4.00 N, F 2 = 6.00 N, and F 3 = 8.00 N are applied to the block, initially at rest on the surface, at angles shown on the diagram. In this problem, you will determine the resultant (total) force vector from the combination of the three individual force vectors. All angles should be measured counterclockwise from the positive x axis (i.e., all angles are positive).  

How far (in meters) will the mass move in 5.0 s? Express the distance d in meters to two significant figures.  

What is the magnitude of the velocity vector of the block at t = 5.0 s? Express your answer in meters per second to two significant figures.

 In what direction is the mass moving at time t = 5.0 s? That is, what angle does the velocity vector make with respect to the positive x axis?

 

Answer 1

m = 2.0 Kg

F1 = 4.0 N
F2 = 6.0 N
F3 = 8.0 N

Fx = F1*cos(25) + F2*cos(360-325) - F3
Fx = 4.0 * cos(25) + 6.0 * cos(35) - 8.0
Fx = 0.54 N

Fy =  F1*sin(25) - F2*sin(360-325)
Fy = 4.0*sin(25) - 6.0 * sin(35)
Fy = -1.75 N

Fnet = sqrt(Fx^2 + Fy^2)
m*a = sqrt(0.54^2 + 1.75^2)
a = 0.916 m/s^2

v = u + a*t
v = 0 + 0.916*5.0
v = 4.58 m/s
Magnitude of Velocity vector, v = 4.58 m/s


Angle = tan^-1(1.75/0.54)
Angle = 72.85^o Clockwise from +ve x axis.