Question below. Please note that previous Chegg answers that I have looked at ar

Question

Question below. Please note that previous Chegg answers that I have looked at are incorrect because they fail to account for the fact that the father is pushing horizontally into the hill. This, of course, will add a 'Y' component to the pushing force which, in turn, will increase the weight force which will [ultimately] increase the normal force. The challenge, then, is in finding the hypotenuse of the father's push (force parallel up the hill which is needed for finding the work). Please help me find the force parallel and then I can solve! Thanks!

A father pushes horizontally on his daughter's sled to move it up a snowy incline, as illustrated in the figure, with h = 3.5 m and ? = 16°. The total mass of the sled and the girl is 35 kg and the coefficient of kinetic friction between the sled runners and the snow is 0.20. If the sled moves up the hill with a constant velocity, how much work is done by the father in moving it from the bottom to the top of the hill?

Explanation / Answer

Start by calculating the normal force:

Fn = mgcos + Fsin = 35kg * 9.8m/s² * cos16º + Fsin16º = 329.71N + 0.27F
Then the friction force Ff = µ*Fn = 0.2*(329.71N + 0.27F) = 65.9N + 0.054F
For "constant velocity", upslope forces = downslope forces
Fcos = mgsin + Ff
Fcos16 = 35kg * 9.8m/s² * sin16 + 65.9N + 0.054F
0.96F = 94.5N + 65.9N + 0.052F
0.90F = 160.4 N
F = 178.22 N
The work done by the force is
work W = Fcos * d = Fcos * h/sin = 178.22N * cos16 * 3.5m / sin16 2175 J

2175 J

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