The diagram below shows a block of mass m=2.00\\; \ m kg on a frictionless horiz
ID: 1855121 • Letter: T
Question
The diagram below shows a block of mass m=2.00; m kg on a frictionless horizontal surface, as seen from above. Three forces of magnitudes F_1 = 4.00;{ m N}, F_2 = 6.00;{ m N}, and F_3 = 8.00;{ m N} 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).
How far (in meters) will the mass move in 5.0 s?
Explanation / Answer
Resultant along X is = F1 cos25 + F2 cos35 -F3 = 0.535 N (+i) ( since cos 325 = cos 35)
Resultant along Y is = F1 sin25 - F2 sin35 = 1.75 N (-j)
Net = 0.5352 + 1.752 = 1.83 N..
= tan-1 (1.75/0.535) = 73 degrees clockwise = 287 degrees anticlock from +X...
a = F/m = 0.915 m/s2
s= 0.5 a t2 = 11.43 m...
velocity direction is same as force..