Self-Assessment Solutions

Forces

1.–176.24lbs
1.4.54kg
1.111.2N
4.460.45N
5.FS = –μSN = –0.3(460.45) = –138.13N and FK = –μKN = –0.25(460.45) = –115.11N
6.Fnet = [369.7 0]
7.The answer is shown in Figure 11.9.
Figure 11.9. The free-body diagram for question 7.

8.Fnet = [88.05 0]

Using Newton's Laws to Determine How Forces Affect an Object's Motion

1.In the same place
2.5m/s @ 45°
3.a = [12.32 0]
4.154m
5.a = [6.05 –10.48]
6.Δr = [27.22 –47.15]

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