The diagram below shows a block of mass m=2.00kg on a frictionless horizontal su
ID: 1275068 • Letter: T
Question
The diagram below shows a block of mass m=2.00kg on a frictionless horizontal surface, as seen from above. Three forces of magnitudes F1=4.00N, F2=6.00N, and F3=8.00N are applied to the block, initially at rest on the surface, at angles shown on the diagram. (Figure 1) 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).
A- Calculate the magnitude of the total resultant force F? r=F? 1+F? 2+F? 3 acting on the mass.
B-What angle does F? r make with the positive x axis?
C-What is the magnitude of the mass's acceleration vector, a? ?
D-What is the direction of a? ? In other words, what angle does this vector make with respect to the positive x axis?
E-How far (in meters) will the mass move in 5.0 s?
F-What is the magnitude of the velocity vector of the block at t=5.0s?
G-In what direction is the mass moving at time t=5.0s? That is, what angle does the velocity vector make with respect to the positive x axis?
Explanation / Answer
Net force in horizontal direction is Fx = F2cos(35) + F1cos()25 -F3
Fx = 0.5401
Net force in vertical direction is Fy = F1sin(25) - F2sin(35)
Fy = -1.751
F = 1.8324
The angle F makes with positive x axis is theta = arccos(F/Fx) =287.14degrees
Acceleration of the mass = F/m = 1.8324/2 = 0.9162m/s^2
2)s =0.5*a*t^2
s= 11.45m
3)v=at = 4.581m/s
4)The direction of velocity is same as that of the Force(or) acceleration. The final velocity makes an angle 287.14 degrees with the positive x axis.