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-Three forces of magnitudes F1=4.0N, F2=6.0N, and F3=8.0N are applied to a block of mass m=2.0kg, initially at rest, at angles shown on the diagram. (Figure 1)In this problem, you will determine the resultant (net) force by combining the three individual force vectors. All angles should be measured counterclockwise from the positive x axis (i.e., all angles are positive).

A) What is the magnitude of the block's acceleration vector, -->a ?
Express your answer to two significant figures.

B) What is the direction of -->a? In other words, what angle does this vector make with respect to the positive x axis?
Express your answer in degrees to two significant figures.

C) How far (in meters) will the block move in 5.0 s? Recall that it starts from rest.
Express the distance d in meters to two significant figures.

oldschool · Answer

) Calculate the magnitude of the resultant force Fx r= -->F1+-->F2+-->F3 acting on the block.

Sum horizontal components:
4*cos25 + 6*cos325 + 8*cos180 = 0.54
Sum vertical components:
4*sin25 + 6*sin325 + 8*sin180 = -1.75
F² = 0.54²+1.75² = 1.8²
A) What is the magnitude of the block's acceleration vector, -->a ?

Express your answer to two significant figures.1.8/2kg = 0.92m/s² A)
B) What is the direction of -->a? In other words, what angle does this vector make with respect to the positive x axis? Express your answer in degrees to two significant figures at arctan(-1.75/0.54) =  287° or about 290°
C) How far (in meters) will the block move in 5.0 s? Recall that it starts from rest.
Express the distance d in meters to two significant figures.
d = ½at² = ½*0.92*5² = 11.5 or about 12m
at arctan(-1.75/0.54) = -73° = 290° B)

az_lender · Answer

Figure 1 has not shown up, so nobody can answer the question.

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