Solve the below parts: Solve the motion of a block with three forces parts D and

Question

Solve the below parts:

Solve the motion of a block with three forces parts D and E.

Your 2.00kg physics book rests on a horizontal tabletop. Find the normal force on the book. Express your answer with the appropriate units. Suppose you pull up on the book with a 15.0-N force. Now what's the normal force? Express your answer with the appropriate units. Suppose you push down with a 15.0-N force. Now what's the normal force? Express your answer with the appropriate units. Three forces of magnitudes F1 = 4.0 N, F2 = 6.0 N, and F3 = 8.0 N are applied to a block of mass to m = 2.0 kg, 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). Express your answer to two significant figures. Significant Figures Feedback: Your answer 0.91 m/s2 was either rounded differently or used a different number of significant figures than required for this part. What is the direction of arightarrow? 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. 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.

Explanation / Answer

4.46 A.

N = 2g = 2 x 9.8 = 19.6 N

B. N = 19.6 - 15 = 4.6 N

C. N = 19.6 + 15 = 34.6 N

Fnet = 4cos25i + 4sin25j - 8i + 6cos35 i - 6sin35j

= 0.54i - 1.75j

|Fnet| = sqrt(0.54^2 + 1.75^2) = 1.83 N

D . angle = 360 - tan-1(1.75 / 0.54) = 287.15 degrees   .........Ans

E .. a =Fnet/m = 1.83 / 2 = 0.915 m/s2

after 5 sec

d = at^2/2 = 0.915 x 5^2/2 = 11.44 m

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