a) The net force on the system is 883.5 N
b) The tension in the rope is 505.5 N
c) The applied force is 1875.8 N
Explanation:
a)
We start by considering the whole system snowmobile+sled. We can apply Newton's second law:
[tex]\sum F = (m+M)a[/tex]
where
[tex]\sum F[/tex] is the net force on the system
m = 150 kg is the mass of the sled
M = 315 kg is the mass of the snowmobile+rider
[tex]a=1.9 m/s^2[/tex] is the acceleration
Substiting the values into the equation, we find the net force:
[tex]\sum F=(150+315)(1.9)=883.5 N[/tex]
b)
Now we consider only the forces acting on the sled. Again, we apply Newton's second law:
[tex]T-F_f = ma[/tex]
where:
T is the tension in the rope, which pulls the sled forward
[tex]F_f[/tex] is the frictional force acting on the sled, which acts backward
m = 150 kg is the mass of the sled
[tex]a=1.9 m/s^2[/tex] is the acceleration
The force of friction on the sled is given by
[tex]F_f = \mu mg[/tex]
where
[tex]\mu=0.15[/tex] is the coefficient of friction of the sled on ice
m = 150 kg
[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity
Substituting and solving for T, we find:
[tex]T-\mu mg = ma\\T=m(a+\mu g)=(150)(1.9+0.15 \cdot 9.8)=505.5 N[/tex]
c)
Now we consider instead the forces acting on the snowmobile+rider only. Applying again Newton's second law,
[tex]F-F_F-T = Ma[/tex]
where
F is the applied force, which pushes the snowmobile forward
[tex]F_F[/tex] is the force of friction acting backward on the snowmobile
T = 505.5 N is the tension in the rope, which pulls the snowmobile backward
M = 315 kg is the mass of the snowmobile+rider
[tex]a=1.9 m/s^2[/tex] is the acceleration
The force of friction on the snowmobile is given by
[tex]F_F = \mu Mg[/tex]
where
[tex]\mu=0.25[/tex] is the coefficient of friction of the snowmobile on ice
m = 315 kg
[tex]g=9.8 m/s^2[/tex]
Substituting and solving for F, we find the applied force:
[tex]F=Ma+\mu Mg+T=(315)(1.9)+(0.25)(315)(9.8)+505.5=1875.8 N[/tex]
Learn more about Newton's second law:
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