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A 72 kg sled is pulled forward from rest by a snowmobile and accelerates at 2.0m/s^2( forward) for 5.0 seconds. The force of friction acting on sled is 120 N (backwards). The total mass of a snowmobile and driver is 450 kg. The drag force acting on snowmobile is 540 N (backwards). Determine the tension in the rope and the force exerted by the snowmobile that pushes the sled forward.

Respuesta :

calculista
  The total force needed to pull the sled is equal  to mass*acceleration plus the force of friction

F_sled = 72kg * 2.0m/s + 120 N = 260 N

Since the tension of the rope must be equal to the force that pulls the sled

tension in the rope = 260N

The force exerted by the snowmobile is equal to the drag force plus its mass times acceleration

F-snowmobile only = 450kg * 2.0m/s + 540N  = 1440 N

F-snowmobile and sled = 1440N  + 260 N = 1700 N

Answer:

Tension force in the rope is 264 N

Force exerted by snowmobile is 1704 N

Explanation:

For sled there is tension force on it in forward direction and friction force on it in backward direction

So here we will have

[tex]F_{net} = ma[/tex]

[tex]F_x - F_f = F_{net}[/tex]

as we know

[tex]F_f = 120 N[/tex]

[tex]T - 120 = 72(2)[/tex]

[tex]T = 120 + 144 = 264 N[/tex]

now for snowmobile and driver we know that their total mass is

m = 450 kg

drag force on it is in backward direction as 540 N

there is tension force on it in backward direction is T = 264 N

now the net force of it is given by

[tex]F - F_d - T = F_{net}[/tex]

[tex]F - 540 - 264 = 450(2)[/tex]

[tex]F = 540 + 264 + 900[/tex]

[tex]F = 1704 N[/tex]

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