WS1
1a. The diagram is the At rest diagram, above
1b. Floor force = Force weight = 85 kg * 10 m/s/s = 850 N
2a. The diagram is the Accelerating upward diagram, above
b. The Fnet= 2.0 m/s/s*85kg = 170 N= F(up) -850 N F(up) = 1020 N
3a. The diagram is the Accelerating downward diagram,above.
3b. The Fnet= 3m/s/s*85kg= 255 N -255N=F(up) -850N F(up) = 595 N
4 The cable breaks,which is the top diagram. F(net) = F(w), because F(up)=0 The scale does not register.
WS2 (We went over thesein class) #1, #2
The Second WS2, #3
When you resolve the triangle, the horizontal portion pulling left = 200N The net force = 200N-75N. Fnet= m*a = 125N= 70kg*a a= 125N/70kg = 1.8m/s/s
WS4, #2
b.Constant speed means Fnet =0, so Fa=Ff= 50N
Mu = 50 N/300 N
c. If friction ended AFTER the object was launched, it’s Net force would be 50 N. 50 N = 30 Kg * a
3. Mu = Ff/FN .15=Ff/500N Ff= .15*500N = 75N
Fnet = Fa + Ff = 100 N – 75N = 25 N 25 N = Fnet = m*a 25N/50 kg = .5m/s/s
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