A72 kg sled is pulled forward from rest by a snowmobile and accelerates at 2 m/s squared forward for five seconds. the force of friction acting on the sled is 120 n backwards. the total mass of the snowmobile and driver is 450 kg. the drag force acting on the snowmobile is 540 n backwards. what is the tension in the rope.

So first make a force diagram. I would label forward forces + and backward forces -. Essentially, the drag force is equal to the Ft (force of tension) + Ff (force of friction on snowmobile - driver system). The force of Friction is equal to mu * Fn. We can find mu through the force of friction acting on the sled. 120 N = mu * Fn (equal to m * g of sled). mu of the Ice is equal to 0.167. So, 540 N = Ft + 0.167 * 4500 N. Ft = -211.5 N. Ft is acting in the backwards direction at a magnitude of 211.5 N

That depends on what kind of environ meant it is in  in space it has the same force that was applied when the velocity was gained