Question

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.

Answer 1

  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 2

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

$F_{net} = ma$

$F_x - F_f = F_{net}$

as we know

$F_f = 120 N$

$T - 120 = 72(2)$

$T = 120 + 144 = 264 N$

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

$F - F_d - T = F_{net}$

$F - 540 - 264 = 450(2)$

$F = 540 + 264 + 900$

$F = 1704 N$