posted by amy on .
jack gives 50 kg jill a ride on a sleigh for 500 meters. he exerts a force of 45 Newtons on the rope of the sled at an angle of 30 degrees. He pulls her at a constant velocity. What is the coefficient of friction between the sled and the ground? How much work does he do on the sled?
ok, im not good at physics but i'll give it a try. this may be a silly question but the ground they are on is snow right?
yes i guess it would be snow
The equation of motion is (in vector form)
vector (m•g) + vector F(fr)+ vector N+vectorF =0
Projections on the axes
x: F•cosα-F(fr) = 0
y: -mg+N+F•sinα = 0
F(fr) = F•cosα,
N =mg- F•sinα,
F(fr) = k•N =k•(mg- F•sinα),
F•cosα= k•(mg- F•sinα),
k = F•cosα/(mg- F•sinα) =
= 45•0.866/(50•9.8-45•0.5) = 0.083.
W= W(F) + W(fr)=F•cosα•s +F(fr) •s=
= F•cosα•s +k•N•s =
= s• (F•cosα + k•(mg - F•sinα)) =
=500 (45•0.866 + 0.083•(50•9.8 - 45•0.5)=38885 J