1.A.
W1=Fscos37°=45⋅(sin37°2)⋅cos37°=119.43 (J) . Answer
1.B.
W2=Fscos180°=mg⋅2⋅cos180°=2⋅9.8⋅2 ⋅cos180°=−39.2 (J) . Answer
1.C.
W3=Ffscos180°=μNscos180°
N=mgcos37+Fsin37°=2⋅9.8⋅cos37+45⋅sin37°=28.65 (N)
W3=μNscos180°=0.2⋅28.65⋅sin37°2⋅cos180°=−19.04 (J) . Answer
1.D.
ΔKE=W1−W2−W3=119.43−39.2−19.04=61.2 (J) . Answer
2.A.
Assume that the angle of inclination of the track is α=50°
F=ma→a=F/m=mgsin50°/m=gsin50°=9.8⋅sin50°=7.5 (m/s2)
v=2sa=2⋅4⋅7.5=7.75 (m/s) . Answer
2.B.
F=ma→a=(F−Ff)/m=(mgsin50°−20)/3=0.84 (m/s2)
v=2sa=2⋅4⋅0.84=2.6 (m/s) . Answer
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