Given info:
Final velocity = vMass of sled = 22.0 kg
Formula used:
F=maF=Forcea=accelerationm=mass
v2=u2+2asv=Final velocityu=initial velocitya=accelerations=distance
Calculation:
Fmax=3f
From x=0 to x=30F=maFmax=22.0 kg*aa=F max22
v2=u2+2asv30= velocity at x=30v302=0+2*F max22*30 mv30= 60 F max 22
From x=30 to x=55Acceleration = a30-55Force = Favg(30-55)Consider, Fmax=3 xForce at x=55 = F55 = -x
Favg(30-55)= F 55− F max2=−f−3f2=−fFavg(30-55) in turms of Fmax, Fmax=3 f F max3=fFavg(30-55)=−xFavg(30-55)=−F max3F=maFavg(30-55)=22.0 kg*aavg(30-55)Favg(30-55)=−F max3−F max3=22.0 kg*aavg(30-55)a=−F max22*3
For this point, u= 60 F max 22v2=u2+2asv55= velocity at x=55v552=60F max22+2*(−F max22*3)*25 mv552=60F max22−50F max22*3
From x=55 to x=90F55−90=Force from x=55 to x=90=-fa55−90=acceleration from x=55 to x=90F55−90 in turms of Fmax,F max3=fF55−90=−F max3F=ma−F max3=22.0 kg*aa55−90=−F max22*3
For this point, u2=60F max22−50F max22*3v2=u2+2asv90= velocity at x=90v552=60F max22−50F max22*3+2*(−F max22*3)*35 mv552=60F max22−50F max22*3−70F max22*3
From x=90 to x=100Acceleration = a90-100Force = Favg(90-100)Favg(90-100)=0−(−f)2=0.5fF max3=fFavg(90-100)=F max6F=maFavg(90-100)=22.0 kg*a90-100Favg(90-100)=F max6F max6=22.0 kg*a90-100a=F max22*6
For this point, u2=60F max22−50F max22*3−70F max22*3v2=u2+2asv100= velocity at x=100=12.5 ms-1v1002=60F max22−50F max22*3−70F max22*3+2*F max22*6*10 mv1002=60F max22−50F max22*3−70F max22*3+20F max22*6=(12.5 ms-1)260F max22−50F max22*3−70F max22*3+20F max22*6=(12.5 ms-1)2156.25=2.73Fmax−0.75Fmax−1.00Fmax+0.15Fmax1.13Fmax=156.25Fmax=138.27 N
Conclusion:
Fmax=138.27 N