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 on the snowmobile is 1875.7 N
Explanation:
a)
To solve this first part, we just analyze all the forces acting in the horizontal direction on the snowmobile+sled system, and we apply Newton's second law, which states that:
[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
Therefore, solving for [tex]\sum F[/tex], we find the net force on the system:
[tex]\sum F = (150+315)(1.9)=883.5 N[/tex]
b)
We can write now the equation of the forces acting on the sled only. We have:
[tex]T-F_f = ma[/tex]
where:
T is the tension in the rope between the sled and the snowmobile, which is pulling the sled forward
[tex]F_f[/tex] is the force of friction acting on the sled
m = 150 kg is the mass of the sled
[tex]a=1.9 m/s^2[/tex] is the acceleration
The force of friction can be written as
[tex]F_f = \mu_k mg[/tex]
where
[tex]\mu_k = 0.15[/tex] is the coefficient of kinetic friction of the sled on ice
Substituting into the previous equation and solving for T, we find the tension:
[tex]T-\mu_k mg = ma\\T=ma+\mu_k mg=(150)(1.9)+(0.15)(150)(9.8)=505.5 N[/tex]
c)
We can now write the equation of the forces acting on the snowmobile, and we have:
[tex]F_a - T - F_F = Ma[/tex]
where:
[tex]F_a[/tex] is the applied force
T = 505.5 N is the tension in the rope, which pulls the snowmobile backward
[tex]F_F[/tex] is the force of friction on the snowmobile
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 can be written as
[tex]F_F = \mu_k Mg[/tex]
where
[tex]\mu_k = 0.25[/tex] is the coefficient of kinetic friction of the snowmobile on ice
Substituting into the previous equation and solving for [tex]F_a[/tex], we find:
[tex]F_a = Ma+T+\mu_k Mg=(315)(1.9)+505.5+(0.25)(315)(9.8)=1875.7 N[/tex]
Learn more about Newton's second law:
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