Question

A father pushes horizontally on his daughter's sled to move it up a snowy incline, as illustrated in the figure, with h = 3.9 m and θ = 11°. 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?

Answer 1

(force exerted by father)+(friction)-weight along the slope = 0

F+(0.2)(35x9.81xcos11)-(35x9.81xsin11)= 0

F= -1.89N (hey, why it is negative? It is wrong. I need more information)

work done = F.S

Answer 2

Force component exerted parallel to slope by man is Fp
Fp = (0.2)(35x9.81xcos11) + (35x9.81xsin11)

(note Fp is up hill, both friction and weight forces are down hill)
Fp = 133 Newtons parallel to slope
motion is parallel to slope
so
work = Fp * hypotenuse length
= 133 * 3.9/sin 11 = 2717 Joules

Answer 3

oh, i know why i am wrong.it is because i take friction a wrong direction. thanks very much!(Damon is correct!)

Answer 4

shouldnt the work = Fp*d*cos(angle between) ?

Answer 5

n9iggaballs

Answer 6

e=mc^2