a father pulls his young daughter on a sled (where the combined mass of his daughter and the sled are 40 kg) with a constant velocity up a snowy hill that is inclined at 14 to the horizontal. a) if the coefficient of kinetic friction between the sled's runners and the snow is 0.15, what is the tension force of the rope on which the father pulls

Explanation:

We are not told what angle the pull rope is to any other reference. That could make a huge difference in the result.

If we ASSUME that the rope pulls parallel to the slope

Then , if T is rope tension

                                  F = ma

T - mgsinθ - μmgcosθ = m(0)

                                 T = mg(sinθ + μcosθ)

                                 T = 40(9.8)(sin14 + 0.15cos14)

                                 T = 151.88677... 150 N

If the rope pulls the sled at a positive angle relative to the slope, As one might expect from an adult holding the rope, then the tension will be increased because only the Tension portion parallel to the slope causes motion in that direction, but will be decreased because the Normal force, and therefore the friction force, of the slope on the sled will be decreased.

from

explanation:

wind is defined by the direction it blows from.   for example, a wind blowing from north to south is called a northerly wind.

the force required to move the cart is divided by three or 33.33 n.